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6.3: Derivatif Separa - Matematik


Setelah kita memeriksa had dan kesinambungan fungsi dua pemboleh ubah, kita dapat melanjutkan kajian derivatif. Ini juga membawa kepada pembezaan.

Derivatif Fungsi Dua Pemboleh ubah

Semasa mengkaji derivatif fungsi satu pemboleh ubah, kami mendapati bahawa satu interpretasi derivatif adalah kadar perubahan seketika (y ) sebagai fungsi (x. ) Notasi Leibniz untuk derivatif adalah (dy / dx , ) yang menunjukkan bahawa (y ) adalah pemboleh ubah bersandar dan (x ) adalah pemboleh ubah tidak bersandar. Untuk fungsi (z = f (x, y) ) dua pemboleh ubah, (x ) dan (y ) adalah pemboleh ubah tidak bersandar dan (z ) adalah pemboleh ubah bersandar. Ini segera menimbulkan dua persoalan: Bagaimana kita menyesuaikan notasi Leibniz untuk fungsi dua pemboleh ubah? Juga, apa tafsiran derivatif? Jawapannya terletak pada terbitan separa.

Definisi: terbitan separa

Mari (f (x, y) ) menjadi fungsi dua pemboleh ubah. Kemudian terbitan separa (f ) berkenaan dengan (x ), ditulis sebagai (∂f / ∂x, ), atau (f_x, ) ditakrifkan sebagai

[ dfrac {∂f} {∂x} = f_x (x, y) = lim_ {h → 0} dfrac {f (x + h, y) −f (x, y)} {h} label {pd1} ]

Derivatif separa (f ) berkenaan dengan (y ), ditulis sebagai (∂f / ∂y ), atau (f_y, ) ditakrifkan sebagai

[ dfrac {∂f} {∂y} = f_y (x, y) = lim_ {k → 0} dfrac {f (x, y + k) −f (x, y)} {k}. label {pd2} ]

Definisi ini menunjukkan dua perbezaan sudah. Pertama, notasi berubah, dalam arti bahawa kita masih menggunakan versi notasi Leibniz, tetapi (d ) dalam notasi asalnya digantikan dengan simbol (∂ ). (Ini bulat ("d" ) biasanya disebut "separa", jadi (∂f / ∂x ) diucapkan sebagai "separa (f ) sehubungan dengan (x ).") Ini adalah petunjuk pertama yang kita hadapi dengan derivatif separa. Kedua, sekarang kita mempunyai dua derivatif berbeza yang dapat kita ambil, kerana terdapat dua pemboleh ubah bebas yang berbeza. Bergantung pada pemboleh ubah mana yang kita pilih, kita dapat menghasilkan derivatif separa yang berbeza sama sekali, dan sering berlaku.

Contoh ( PageIndex {1} ): Mengira Derivatif Separa dari Definisi

Gunakan definisi terbitan separa sebagai had untuk mengira (∂f / ∂x ) dan (∂f / ∂y ) untuk fungsi

[f (x, y) = x ^ 2−3xy + 2y ^ 2−4x + 5y − 12. nombor ]

Penyelesaian

Pertama, hitung (f (x + h, y). )

[ start {align *} f (x + h, y) & = (x + h) ^ 2−3 (x + h) y + 2y ^ 2−4 (x + h) + 5y − 12 bukan angka & = x ^ 2 + 2xh + h ^ 2−3xy − 3hy + 2y ^ 2−4x − 4h + 5y − 12. end {align *} ]

Seterusnya, gantikan ini ke Persamaan ref {pd1} dan permudahkan:

[ start {align *} dfrac {∂f} {∂x} & = lim_ {h → 0} dfrac {f (x + h, y) −f (x, y)} {h} bukan nombor & = lim_ {h → 0} dfrac {(x ^ 2 + 2xh + h ^ 2−3xy − 3hy + 2y ^ 2−4x − 4h + 5y − 12) - (x ^ 2−3xy + 2y ^ 2−4x + 5y − 12)} {h} nonumber & = lim_ {h → 0} dfrac {x ^ 2 + 2xh + h ^ 2−3xy − 3hy + 2y ^ 2−4x− 4h + 5y − 12 − x ^ 2 + 3xy − 2y ^ 2 + 4x − 5y + 12} {h} nonumber & = lim_ {h → 0} dfrac {2xh + h ^ 2−3hy − 4h } {h} nonumber & = lim_ {h → 0} dfrac {h (2x + h − 3y − 4)} {h} nonumber & = lim_ {h → 0} (2x + h − 3y − 4) bukan nombor & = 2x − 3y − 4. end {align *} ]

Untuk mengira ( dfrac {∂f} {∂y} ), hitung terlebih dahulu (f (x, y + h): )

[ start {align *} f (x + h, y) & = x ^ 2−3x (y + h) +2 (y + h) ^ 2−4x + 5 (y + h) −12 bukan angka & = x ^ 2−3xy − 3xh + 2y ^ 2 + 4yh + 2h ^ 2−4x + 5y + 5h − 12. end {align *} ]

Seterusnya, gantikan ini ke Persamaan ref {pd2} dan permudahkan:

[ start {align *} dfrac {∂f} {∂y} & = lim_ {h → 0} dfrac {f (x, y + h) −f (x, y)} {h} bukan nombor & = lim_ {h → 0} dfrac {(x ^ 2−3xy − 3xh + 2y ^ 2 + 4yh + 2h ^ 2−4x + 5y + 5h − 12) - (x ^ 2−3xy + 2y ^ 2−4x + 5y − 12)} {h} nonumber & = lim_ {h → 0} dfrac {x ^ 2−3xy − 3xh + 2y ^ 2 + 4yh + 2h ^ 2−4x + 5y + 5h − 12 − x ^ 2 + 3xy − 2y ^ 2 + 4x − 5y + 12} {h} nonumber & = lim_ {h → 0} dfrac {−3xh + 4yh + 2h ^ 2 + 5h} {h} nonumber & = lim_ {h → 0} dfrac {h (−3x + 4y + 2h + 5)} {h} nonumber & = lim_ {h → 0} ( −3x + 4y + 2h + 5) nonumber & = - 3x + 4y + 5 end {align *} ]

Latihan ( PageIndex {1} )

Gunakan definisi terbitan separa sebagai had untuk mengira (∂f / ∂x ) dan (∂f / ∂y ) untuk fungsi

[f (x, y) = 4x ^ 2 + 2xy − y ^ 2 + 3x − 2y + 5. nonumber ]

Petunjuk

Gunakan Persamaan ref {pd1} dan ref {pd2} dari definisi terbitan separa.

Jawapan

( dfrac {∂f} {∂x} = 8x + 2y + 3 )

( dfrac {∂f} {∂y} = 2x − 2y − 2 )

Idea yang perlu diingat semasa mengira terbitan separa adalah untuk memperlakukan semua pemboleh ubah bebas, selain pemboleh ubah berkenaan dengan yang kita bezakan, sebagai pemalar. Kemudian teruskan untuk membezakan seperti fungsi satu pemboleh ubah. Untuk melihat mengapa ini benar, betulkan dahulu (y ) dan tentukan (g (x) = f (x, y) ) sebagai fungsi (x ). Kemudian

[ start {align} g ′ (x) & = lim_ {h → 0} dfrac {g (x + h) −g (x)} {h} nonumber [6pt] & = lim_ {h → 0} dfrac {f (x + h, y) −f (x, y)} {h} nonumber [6pt] & = dfrac {∂f} {∂x}. nonumber end {align} ]

Perkara yang sama berlaku untuk mengira terbitan separa (f ) berkenaan dengan (y ). Kali ini, betulkan (x ) dan tentukan (h (y) = f (x, y) ) sebagai fungsi (y ). Kemudian

[ start {align} h ′ (x) & = lim_ {k → 0} dfrac {h (x + k) −h (x)} {k} nonumber [6pt] & = lim_ {k → 0} dfrac {f (x, y + k) −f (x, y)} {k} nonumber [6pt] & = dfrac {∂f} {∂y}. nonumber end {align} ]

Semua peraturan pembezaan berlaku.

Contoh ( PageIndex {2} ): Mengira Derivatif Separa

Hitungkan (∂f / ∂x ) dan (∂f / ∂y ) untuk fungsi berikut dengan menahan pemalar pemboleh ubah yang berlawanan kemudian membezakan:

  1. (f (x, y) = x ^ 2−3xy + 2y ^ 2−4x + 5y − 12 )
  2. (g (x, y) = sin (x ^ 2y − 2x + 4) )

Penyelesaian:

a. Untuk mengira (∂f / ∂x ), perlakukan pemboleh ubah (y ) sebagai pemalar. Kemudian bezakan (f (x, y) ) berkenaan dengan (x ) menggunakan jumlah, perbezaan, dan peraturan kuasa:

[ start {align *} dfrac {∂f} {∂x} & = dfrac {∂} {∂x} kiri [x ^ 2−3xy + 2y ^ 2−4x + 5y − 12 kanan] nonumber [6pt] & = dfrac {∂} {∂x} [x ^ 2] - dfrac {∂} {∂x} [3xy] + dfrac {∂} {∂x} [2y ^ 2 ] - dfrac {∂} {∂x} [4x] + dfrac {∂} {∂x} [5y] - dfrac {∂} {∂x} [12] nonumber [6pt] & = 2x −3y + 0−4 + 0−0 nonumber & = 2x − 3y − 4. nonumber end {align *} ]

Derivatif dari istilah ketiga, kelima, dan keenam semuanya sifar kerana tidak mengandungi pemboleh ubah (x ), jadi mereka dianggap sebagai istilah tetap. Derivatif dari istilah kedua sama dengan pekali (x ), yang (- 3y ). Mengira (∂f / ∂y ):

[ start {align *} dfrac {∂f} {∂y} & = dfrac {∂} {∂y} kiri [x ^ 2−3xy + 2y ^ 2−4x + 5y − 12 kanan] nonumber [6pt] & = dfrac {∂} {∂y} [x ^ 2] - dfrac {∂} {∂y} [3xy] + dfrac {∂} {∂y} [2y ^ 2 ] - dfrac {∂} {∂y} [4x] + dfrac {∂} {∂y} [5y] - dfrac {∂} {∂y} [12] nonumber [6pt] & = - 3x + 4y − 0 + 5−0 nonumber & = - 3x + 4y + 5. nonumber end {align *} ]

Ini adalah jawapan yang sama yang diperoleh dalam Contoh ( PageIndex {1} ).

b. Untuk mengira (∂g / ∂x, ) perlakukan pemboleh ubah y sebagai pemalar. Kemudian bezakan (g (x, y) ) berkenaan dengan (x ) menggunakan peraturan rantai dan peraturan kuasa:

[ start {align *} dfrac {∂g} {∂x} & = dfrac {∂} {∂x} kiri [ sin (x ^ 2y − 2x + 4) kanan] bukan nombor [6pt] & = cos (x ^ 2y − 2x + 4) dfrac {∂} {∂x} [x ^ 2y − 2x + 4] nonumber [6pt] & = (2xy − 2) cos (x ^ 2y − 2x + 4). nonumber end {align *} ]

Untuk mengira (∂g / ∂y, ) perlakukan pemboleh ubah (x ) sebagai pemalar. Kemudian bezakan (g (x, y) ) berkenaan dengan (y ) menggunakan peraturan rantai dan peraturan kuasa:

[ start {align *} dfrac {∂g} {∂y} & = dfrac {∂} {∂y} kiri [ sin (x ^ 2y − 2x + 4) kanan] bukan nombor [6pt] & = cos (x ^ 2y − 2x + 4) dfrac {∂} {∂y} [x ^ 2y − 2x + 4] nonumber [6pt] & = x ^ 2 cos (x ^ 2y − 2x + 4). nonumber end {align *} ]

Latihan ( PageIndex {2} )

Hitungkan (∂f / ∂x ) dan (∂f / ∂y ) untuk fungsi

[f (x, y) = tan (x ^ 3−3x ^ 2y ^ 2 + 2y ^ 4) bukan nombor ]

dengan menahan pemalar pemboleh ubah yang berlawanan, kemudian membezakan.

Petunjuk

Gunakan Persamaan ref {pd1} dan ref {pd1} dari definisi terbitan separa.

Jawapan

( dfrac {∂f} {∂x} = (3x ^ 2−6xy ^ 2) sec ^ 2 (x ^ 3−3x ^ 2y ^ 2 + 2y ^ 4) )

( dfrac {∂f} {∂y} = (- 6x ^ 2y + 8y ^ 3) sec ^ 2 (x ^ 3−3x ^ 2y ^ 2 + 2y ^ 4) )

Bagaimana kita boleh mentafsirkan derivatif separa ini? Ingat bahawa graf fungsi dua pemboleh ubah adalah permukaan di (R ^ 3 ). Sekiranya kita membuang had dari definisi terbitan separa sehubungan dengan (x ), hasil perbezaan tetap:

[ dfrac {f (x + h, y) −f (x, y)} {h}. ]

Ini menyerupai hasil bagi perbezaan dari satu fungsi satu pemboleh ubah, kecuali adanya pemboleh ubah (y ). Rajah ( PageIndex {1} ) menggambarkan permukaan yang dijelaskan oleh fungsi sewenang-wenangnya (z = f (x, y). )

Dalam Rajah ( PageIndex {1} ), nilai (h ) adalah positif. Sekiranya kita graf (f (x, y) ) dan (f (x + h, y) ) untuk titik sewenang-wenang ((x, y), ) maka cerun garis pemisah yang melewati ini dua mata diberikan oleh

[ dfrac {f (x + h, y) −f (x, y)} {h}. ]

Garis ini selari dengan paksi (x ) -. Oleh itu, cerun garis pemisah mewakili kadar perubahan fungsi rata-rata (f ) ketika kita bergerak selari dengan paksi (x ) -. Apabila (h ) menghampiri sifar, cerun garis pemisah menghampiri cerun garis tangen.

Sekiranya kita memilih untuk menukar (y ) dan bukan (x ) dengan nilai kenaikan yang sama (h ), maka garis pemisah selari dengan paksi (y ) - dan begitu juga dengan garis tangen. Oleh itu, (∂f / ∂x ) mewakili cerun garis tangen yang melewati titik ((x, y, f (x, y)) ) selari dengan paksi (x ) - dan (∂f / ∂y ) mewakili cerun garis tangen yang melewati titik ((x, y, f (x, y)) ) selari dengan paksi (y ) -. Sekiranya kita ingin mencari cerun garis tangen yang melewati titik yang sama ke arah lain, maka kita memerlukan apa yang disebut turunan arah.

Kami sekarang kembali ke idea peta kontur, yang kami perkenalkan dalam Fungsi Beberapa Pembolehubah. Kita boleh menggunakan a peta kontur untuk menganggarkan terbitan separa fungsi (g (x, y) ).

Contoh ( PageIndex {3} ): Derivatif Separa dari Peta Kontur

Gunakan peta kontur untuk mengira (∂g / ∂x ) pada titik (( sqrt {5}, 0) ) untuk fungsi

[g (x, y) = sqrt {9 − x ^ 2 − y ^ 2}. nombor ]

Penyelesaian

Rajah ( PageIndex {2} ) mewakili peta kontur untuk fungsi (g (x, y) ).

Lingkaran dalaman pada peta kontur sepadan dengan (c = 2 ) dan bulatan seterusnya sesuai dengan (c = 1 ). Lingkaran pertama diberikan oleh persamaan (2 = sqrt {9 − x ^ 2 − y ^ 2} ); bulatan kedua diberikan oleh persamaan (1 = sqrt {9 − x ^ 2 − y ^ 2} ). Persamaan pertama menyederhanakan menjadi (x ^ 2 + y ^ 2 = 5 ) dan persamaan kedua menyederhanakan menjadi (x ^ 2 + y ^ 2 = 8. ) The (x ) - pintasan bulatan pertama ialah (( sqrt {5}, 0) ) dan (x ) - pintasan bulatan kedua adalah ((2 sqrt {2}, 0) ). Kita dapat menganggar nilai (∂g / ∂x ) yang dinilai pada titik (( sqrt {5}, 0) ) menggunakan formula cerun:

[ start {align *} kiri. dfrac {∂g} {∂x} kanan | _ {(x, y) = ( sqrt {5}, 0)} ≈ dfrac {g ( sqrt {5}, 0) −g (2 sqrt {2}, 0)} { sqrt {5} −2 sqrt {2}} = dfrac {2−1} { sqrt {5} −2 sqrt {2}} = dfrac {1} { sqrt {5} −2 sqrt {2}} ≈ − 1.688. end {align *} ]

Untuk mengira nilai tepat (∂g / ∂x ) yang dinilai pada titik (( sqrt {5}, 0) ), kita mulakan dengan mencari (∂g / ∂x ) menggunakan peraturan rantai . Pertama, kita menulis semula fungsi sebagai

[g (x, y) = sqrt {9 − x ^ 2 − y ^ 2} = (9 − x ^ 2 − y ^ 2) ^ {1/2} ]

dan kemudian bezakan berkenaan dengan (x ) sambil menahan pemalar (y ):

[ start {align *} dfrac {∂g} {∂x} & = dfrac {1} {2} (9 − x ^ 2 − y ^ 2) ^ {- 1/2} (- 2x) [5pt] & = - dfrac {x} { sqrt {9 − x ^ 2 − y ^ 2}}. end {align *} ]

Seterusnya, kami menilai ungkapan ini menggunakan (x = sqrt {5} ) dan (y = 0 ):

[ dfrac {∂g} {∂x} ∣ _ {(x, y) = ( sqrt {5}, 0)} = - dfrac { sqrt {5}} { sqrt {9 - ( sqrt {5}) ^ 2− (0) ^ 2}} = - dfrac { sqrt {5}} { sqrt {4}} = - dfrac { sqrt {5}} {2} ≈ − 1.118 . ]

Anggaran untuk derivatif separa sepadan dengan cerun garis pemisah yang melewati titik (( sqrt {5}, 0, g ( sqrt {5}, 0)) ) dan ((2 sqrt { 2}, 0, g (2 sqrt {2}, 0)) ). Ini mewakili penghampiran ke lereng garis singgung ke permukaan melalui titik (( sqrt {5}, 0, g ( sqrt {5}, 0)), ) yang selari dengan (x ) - paksi.

Latihan ( PageIndex {3} )

Gunakan peta kontur untuk menganggarkan (∂f / ∂y ) pada titik ((0, sqrt {2}) ) untuk fungsi

[f (x, y) = x ^ 2 − y ^ 2. bukan nombor ]

Bandingkan dengan jawapan yang tepat.

Petunjuk

Buat peta kontur untuk (f ) menggunakan nilai (c ) dari (- 3 ) hingga (3 ). Lengkung berikut yang manakah melalui titik ((0, sqrt {2})? )

Jawapan

Dengan menggunakan lekukan yang sesuai dengan (c = −2 ) dan (c = −3, ), kami memperoleh

[ dibiarkan. dfrac {∂f} {∂y} right∣ _ {(x, y) = (0, sqrt {2})} ≈ dfrac {f (0, sqrt {3}) - f (0, sqrt {2})} { sqrt {3} - sqrt {2}} = dfrac {−3 - (- 2)} { sqrt {3} - sqrt {2}} ⋅ dfrac { sqrt {3} + sqrt {2}} { sqrt {3} + sqrt {2}} = - sqrt {3} - sqrt {2} − − 3.146. nombor ]

Jawapan yang tepat adalah

[ dibiarkan. dfrac {∂f} {∂y} kanan | _ {(x, y) = (0, sqrt {2})} = (- 2y | _ {(x, y) = (0, sqrt { 2})} = - 2 sqrt {2} ≈ − 2.828. Bukan nombor ]

Fungsi Lebih Dari Dua Pembolehubah

Katakan kita mempunyai fungsi dari tiga pemboleh ubah, seperti (w = f (x, y, z). ) Kita dapat mengira turunan separa dari (w ) berkenaan dengan mana-mana pemboleh ubah bebas, hanya sebagai peluasan definisi bagi sebahagian terbitan fungsi dua pemboleh ubah.

Definisi: Derivatif Separa

Mari (f (x, y, z) ) menjadi fungsi tiga pemboleh ubah. Kemudian, terbitan separa (f ) berkenaan dengan (x ), ditulis sebagai (∂f / ∂x, ) atau (f_x, ) ditakrifkan sebagai

[ dfrac {∂f} {∂x} = f_x (x, y, z) = lim_ {h → 0} dfrac {f (x + h, y, z) −f (x, y, z )} {h}. label {PD2a} ]

The terbitan separa daripada (f ) dengan hormat untuk (y ), ditulis sebagai (∂f / ∂y ), atau (f_y ), ditakrifkan sebagai

[ dfrac {∂f} {∂y} = f_y (x, y, z) = lim_ {k → 0} dfrac {f (x, y + k, z) −f (x, y, z )} {k.} label {PD2b} ]

The terbitan separa (f ) sehubungan dengan (z ), ditulis sebagai (∂f / ∂z ), atau (f_z ), didefinisikan sebagai

[ dfrac {∂f} {∂z} = f_z (x, y, z) = lim_ {m → 0} dfrac {f (x, y, z + m) −f (x, y, z )} {m}. label {PD2c} ]

Kita dapat mengira turunan separa dari fungsi tiga pemboleh ubah menggunakan idea yang sama yang kita gunakan untuk fungsi dua pemboleh ubah. Sebagai contoh, jika kita mempunyai fungsi (f ) dari (x, y ), dan (z ), dan kita ingin mengira (∂f / ∂x ), maka kita memperlakukan dua yang lain pemboleh ubah bebas seolah-olah ia adalah pemalar, kemudian bezakan berkenaan dengan (x ).

Contoh ( PageIndex {4} ): Mengira Derivatif Separa untuk Fungsi Tiga Pembolehubah

Gunakan definisi had derivatif separa untuk mengira (∂f / ∂x ) untuk fungsi tersebut

[f (x, y, z) = x ^ 2−3xy + 2y ^ 2−4xz + 5yz ^ 2−12x + 4y − 3z. nombor ]

Kemudian, cari (∂f / ∂y ) dan (∂f / ∂z ) dengan menetapkan dua pemboleh ubah lain yang tetap dan membezakan dengan sewajarnya.

Penyelesaian:

Kami mula-mula mengira (∂f / ∂x ) menggunakan Persamaan ref {PD2a}, kemudian kami mengira dua terbitan separa yang lain dengan menahan pemboleh ubah yang selebihnya tetap. Untuk menggunakan persamaan untuk mencari (∂f / ∂x ), pertama kita perlu mengira (f (x + h, y, z): )

[ mula {align *} f (x + h, y, z) & = (x + h) ^ 2−3 (x + h) y + 2y ^ 2−4 (x + h) z + 5yz ^ 2−12 (x + h) + 4y − 3z nonumber & = x ^ 2 + 2xh + h ^ 2−3xy − 3xh + 2y ^ 2−4xz − 4hz + 5yz ^ 2−12x − 12h + 4y− 3z nonumber end {align *} ]

dan ingat bahawa (f (x, y, z) = x ^ 2−3xy + 2y ^ 2−4zx + 5yz ^ 2−12x + 4y − 3z. ) Seterusnya, kami menggantikan kedua ungkapan ini ke dalam persamaan:

[ start {align *} dfrac {∂f} {∂x} & = lim_ {h → 0} kiri [ dfrac {x ^ 2 + 2xh + h ^ 2−3xy − 3hy + 2y ^ 2 −4xz − 4hz + 5yz ^ 2−12x − 12h + 4y − 3zh − x ^ 2−3xy + 2y ^ 2−4xz + 5yz ^ 2−12x + 4y − 3z} {h} kanan] bukan nombor & = lim_ {h → 0} kiri [ dfrac {2xh + h ^ 2−3hy − 4hz − 12h} {h} kanan] nonumber & = lim_ {h → 0} kiri [ dfrac {h (2x + h − 3y − 4z − 12)} {h} kanan] bukan nombor & = lim_ {h → 0} (2x + h − 3y − 4z − 12) bukan nombor & = 2x − 3y − 4z − 12. end {align *} ]

Kemudian kita dapati (∂f / ∂y ) dengan menahan pemalar (x ) dan (z ). Oleh itu, sebarang istilah yang tidak termasuk pemboleh ubah (y ) adalah tetap, dan terbitannya adalah sifar. Kita dapat menerapkan jumlah, perbezaan, dan peraturan daya untuk fungsi satu pemboleh ubah:

[ start {align *} & dfrac {∂} {∂y} kiri [x ^ 2−3xy + 2y ^ 2−4xz + 5yz ^ 2−12x + 4y − 3z kanan] nonumber & = dfrac {∂} {∂y} [x ^ 2] - dfrac {∂} {∂y} [3xy] + dfrac {∂} {∂y} [2y ^ 2] - dfrac {∂} { ∂y} [4xz] + dfrac {∂} {∂y} [5yz ^ 2] - dfrac {∂} {∂y} [12x] + dfrac {∂} {∂y} [4y] - dfrac {∂} {∂z} [3z] nonumber & = 0−3x + 4y − 0 + 5z ^ 2−0 + 4−0 nonumber & = - 3x + 4y + 5z ^ 2 + 4. end {align *} ]

Untuk mengira (∂f / ∂z, ), kami menahan pemalar (x ) dan (y ) dan menerapkan jumlah, perbezaan, dan peraturan kuasa untuk fungsi satu pemboleh ubah:

[ start {align *} & dfrac {∂} {∂z} [x ^ 2−3xy + 2y ^ 2−4xz + 5yz ^ 2−12x + 4y − 3z] nonumber & = dfrac { ∂} {∂z} [x ^ 2] - dfrac {∂} {∂z} [3xy] + dfrac {∂} {∂z} [2y ^ 2] - dfrac {∂} {∂z} [ 4xz] + dfrac {∂} {∂z} [5yz ^ 2] - dfrac {∂} {∂z} [12x] + dfrac {∂} {∂z} [4y] - dfrac {∂} { ∂z} [3z] nonumber & = 0−0 + 0−4x + 10yz − 0 + 0−3 nonumber & = - 4x + 10yz − 3 end {align *} ]

Latihan ( PageIndex {4} )

Gunakan definisi had derivatif separa untuk mengira (∂f / ∂x ) untuk fungsi tersebut

[f (x, y, z) = 2x ^ 2−4x ^ 2y + 2y ^ 2 + 5xz ^ 2−6x + 3z − 8. nonumber ]

Kemudian cari (∂f / ∂y ) dan (∂f / ∂z ) dengan menetapkan dua pemboleh ubah lain yang tetap dan membezakan dengan sewajarnya.

Petunjuk

Gunakan strategi dalam contoh sebelumnya.

Jawapan

( dfrac {∂f} {∂x} = 4x − 8xy + 5z ^ 2−6, dfrac {∂f} {∂y} = - 4x ^ 2 + 4y, dfrac {∂f} {∂z } = 10xz + 3 )

Contoh ( PageIndex {5} ): Mengira Derivatif Separa untuk Fungsi Tiga Pembolehubah

Hitung tiga terbitan separa fungsi berikut.

  1. (f (x, y, z) = x ^ 2y − 4xz + y ^ 2x − 3yz )
  2. (g (x, y, z) = sin (x ^ 2y − z) + cos (x ^ 2 − yz) )

Penyelesaian

Dalam setiap kes, perlakukan semua pemboleh ubah sebagai pemalar kecuali yang terbitan separa yang anda hitung.

a.

[ start {align *} dfrac {∂f} {∂x} & = dfrac {∂} {∂x} kiri [ dfrac {x ^ 2y − 4xz + y ^ 2} {x − 3yz} kanan] nonumber [6pt] & = dfrac { dfrac {∂} {∂x} (x ^ 2y − 4xz + y ^ 2) (x − 3yz) - (x ^ 2y − 4xz + y ^ 2) dfrac {∂} {∂x} (x − 3yz)} {(x − 3yz) ^ 2} nonumber [6pt] & = dfrac {(2xy − 4z) (x − 3yz) - ( x ^ 2y − 4xz + y ^ 2) (1)} {(x − 3yz) ^ 2} nonumber [6pt] & = dfrac {2x ^ 2y − 6xy ^ 2z − 4xz + 12yz ^ 2 − x ^ 2y + 4xz − y ^ 2} {(x − 3yz) ^ 2} nonumber [6pt] & = dfrac {x ^ 2y − 6xy ^ 2z − 4xz + 12yz ^ 2 + 4xz − y ^ 2} {(x − 3yz) ^ 2} nonumber end {align *} ]

[ start {align *} dfrac {∂f} {∂y} & = dfrac {∂} {∂y} kiri [ dfrac {x ^ 2y − 4xz + y ^ 2} {x − 3yz} kanan] nonumber [6pt] & = dfrac { dfrac {∂} {∂y} (x ^ 2y − 4xz + y ^ 2) (x − 3yz) - (x ^ 2y − 4xz + y ^ 2) dfrac {∂} {∂y} (x − 3yz)} {(x − 3yz) ^ 2} nonumber [6pt] & = dfrac {(x ^ 2 + 2y) (x − 3yz) - (x ^ 2y − 4xz + y ^ 2) (- 3z)} {(x − 3yz) ^ 2} nonumber [6pt] & = dfrac {x ^ 3−3x ^ 2yz + 2xy − 6y ^ 2z + 3x ^ 2yz − 12xz ^ 2 + 3y ^ 2z} {(x − 3yz) ^ 2} nonumber [6pt] & = dfrac {x ^ 3 + 2xy − 3y ^ 2z − 12xz ^ 2} { (x − 3yz) ^ 2} nonumber end {align *} ]

[ start {align *} dfrac {∂f} {∂z} & = dfrac {∂} {∂z} kiri [ dfrac {x ^ 2y − 4xz + y ^ 2} {x − 3yz} kanan] nonumber [6pt] & = dfrac { dfrac {∂} {∂z} (x ^ 2y − 4xz + y ^ 2) (x − 3yz) - (x ^ 2y − 4xz + y ^ 2) dfrac {∂} {∂z} (x − 3yz)} {(x − 3yz) ^ 2} nonumber [6pt] & = dfrac {(- 4x) (x − 3yz) - (x ^ 2y − 4xz + y ^ 2) (- 3y)} {(x − 3yz) ^ 2} nonumber [6pt] & = dfrac {−4x ^ 2 + 12xyz + 3x ^ 2y ^ 2−12xyz + 3y ^ 3} {(x − 3yz) ^ 2} nonumber [6pt] & = dfrac {−4x ^ 2 + 3x ^ 2y ^ 2 + 3y ^ 3} {(x − 3yz) ^ 2} bukan nombor end {align *} ]

b.

[ start {align *} dfrac {∂f} {∂x} & = dfrac {∂} {∂x} kiri [ sin (x ^ 2y − z) + cos (x ^ 2 − yz ) kanan] nonumber [6pt] & = ( cos (x ^ 2y − z)) dfrac {∂} {∂x} (x ^ 2y − z) - ( sin (x ^ 2 − yz) )) dfrac {∂} {∂x} (x ^ 2 − yz) nonumber [6pt] & = 2xy cos (x ^ 2y − z) −2x sin (x ^ 2 − yz) angka end {align *} ]

[ start {align *} dfrac {∂f} {∂y} & = dfrac {∂} {∂y} [ sin (x ^ 2y − z) + cos (x ^ 2 − yz)] nonumber [6pt] & = ( cos (x ^ 2y − z)) dfrac {∂} {∂y} (x ^ 2y − z) - ( sin (x ^ 2 − yz)) dfrac {∂} {∂y} (x ^ 2 − yz) nonumber [6pt] & = x ^ 2 cos (x ^ 2y − z) + z sin (x ^ 2 − yz) nonumber end {sejajar *} ]

[ start {align *} dfrac {∂f} {∂z} & = dfrac {∂} {∂z} [ sin (x ^ 2y − z) + cos (x ^ 2 − yz)] nonumber [6pt] & = ( cos (x ^ 2y − z)) dfrac {∂} {∂z} (x ^ 2y − z) - ( sin (x ^ 2 − yz)) dfrac {∂} {∂z} (x ^ 2 − yz) nonumber [6pt] & = - cos (x ^ 2y − z) + y sin (x ^ 2 − yz) nonumber end {align *} ]

Latihan ( PageIndex {5} )

Hitungkan (∂f / ∂x, ∂f / ∂y, ) dan (∂f / ∂z ) untuk fungsi

[f (x, y, z) = sec (x ^ 2y) - tan (x ^ 3yz ^ 2). nombor ]

Petunjuk

Gunakan strategi dalam contoh sebelumnya.

Jawapan

( dfrac {∂f} {∂x} = 2xy sec (x ^ 2y) tan (x ^ 2y) −3x ^ 2yz ^ 2 sec ^ 2 (x ^ 3yz ^ 2) )

( dfrac {∂f} {∂y} = x ^ 2 sec (x ^ 2y) tan (x ^ 2y) −x ^ 3z ^ 2 sec ^ 2 (x ^ 3yz ^ 2) )

( dfrac {∂f} {∂z} = - 2x ^ 3yz sec ^ 2 (x ^ 3yz ^ 2) )

Derivatif Separa Tertinggi

Pertimbangkan fungsinya

[f (x, y) = 2x ^ 3−4xy ^ 2 + 5y ^ 3−6xy + 5x − 4y + 12. ]

Derivatif separa adalah

[ dfrac {∂f} {∂x} = 6x ^ 2−4y ^ 2−6y + 5 ]

dan

[ dfrac {∂f} {∂y} = - 8xy + 15y ^ 2−6x − 4. ]

Setiap derivatif separa ini adalah fungsi dari dua pemboleh ubah, jadi kita dapat mengira terbitan separa dari fungsi-fungsi ini. Sama seperti derivatif fungsi pemboleh ubah tunggal, kita dapat memanggilnya derivatif pesanan kedua, derivatif pesanan ketiga, dan sebagainya. Secara umum, mereka disebut sebagai derivatif separa tertib yang lebih tinggi. Terdapat empat derivatif separa pesanan kedua untuk sebarang fungsi (dengan syarat semuanya ada):

[ start {align *} dfrac {∂ ^ 2f} {∂x ^ 2} & = dfrac {∂} {∂x} kiri [ dfrac {∂f} {∂x} kanan] [5pt] dfrac {∂ ^ 2f} {∂x , ∂y} & = dfrac {∂} {∂y} kiri [ dfrac {∂f} {∂x} kanan] [5pt] dfrac {∂ ^ 2f} {∂y , ∂x} & = dfrac {∂} {∂x} kiri [ dfrac {∂f} {∂y} kanan] [5pt] dfrac { ∂ ^ 2f} {∂y ^ 2} & = dfrac {∂} {∂y} kiri [ dfrac {∂f} {∂y} kanan]. End {align *} ]

Notasi alternatif untuk masing-masing adalah (f_ {xx}, f_ {xy}, f_ {yx}, ) dan (f_ {yy} ). Derivatif separa tertib tinggi yang dikira berkenaan dengan pemboleh ubah yang berbeza, seperti (f_ {xy} ) dan (f_ {yx} ), biasanya disebut terbitan separa campuran.

Contoh ( PageIndex {6} ): Mengira Derivatif Separa Kedua

Hitung keempat-empat derivatif separa kedua untuk fungsi tersebut

[f (x, y) = xe ^ {- 3y} + sin (2x − 5y). label {Ex6e1} ]

Penyelesaian:

Untuk mengira ( dfrac {∂ ^ 2f} {∂x ^ 2} ) dan ( dfrac {∂ ^ 2f} {∂x∂y} ), pertama-tama kita mengira (∂f / ∂x ) :

[ dfrac {∂f} {∂x} = e ^ {- 3y} +2 cos (2x − 5y). label {Ex6e2} ]

Untuk mengira ( dfrac {∂ ^ 2f} {∂x ^ 2} ), bezakan (∂f / ∂x ) (Persamaan ref {Ex6e2}) berkenaan dengan (x ):

[ start {align *} dfrac {∂ ^ 2f} {∂x ^ 2} & = dfrac {∂} {∂x} kiri [ dfrac {∂f} {∂x} kanan] bukan nombor [6pt] & = dfrac {∂} {∂x} [e ^ {- 3y} +2 cos (2x − 5y)] nonumber [6pt] & = - 4 sin (2x − 5y ). nonumber end {align *} nonumber ]

Untuk mengira ( dfrac {∂ ^ 2f} {∂x∂y} ), bezakan (∂f / ∂x ) (Persamaan ref {Ex6e2}) berkenaan dengan (y ):

[ start {align *} dfrac {∂ ^ 2f} {∂x , ∂y} & = dfrac {∂} {∂y} kiri [ dfrac {∂f} {∂x} kanan] nonumber [6pt] & = dfrac {∂} {∂y} [e ^ {- 3y} +2 cos (2x − 5y)] nonumber [6pt] & = - 3e ^ {- 3y } +10 sin (2x − 5y). nonumber end {align *} nonumber ]

Untuk mengira ( dfrac {∂ ^ 2f} {∂x∂y} ) dan ( dfrac {∂ ^ 2f} {∂y ^ 2} ), hitung terlebih dahulu (∂f / ∂y ):

[ dfrac {∂f} {∂y} = - 3xe ^ {- 3y} −5 cos (2x − 5y). label {Ex6e5} ]

Untuk mengira ( dfrac {∂ ^ 2f} {∂y∂x} ), bezakan (∂f / ∂y ) (Persamaan ref {Ex6e5}) berkenaan dengan (x ):

[ start {align *} dfrac {∂ ^ 2f} {∂y , ∂x} & = dfrac {∂} {∂x} kiri [ dfrac {∂f} {∂y} kanan] nonumber [6pt] & = dfrac {∂} {∂x} [- 3xe ^ {- 3y} −5 cos (2x − 5y)] nonumber [6pt] & = - 3e ^ {- 3y} +10 sin (2x − 5y). nonumber end {align *} nonumber ]

Untuk mengira ( dfrac {∂ ^ 2f} {∂y ^ 2} ), bezakan (∂f / ∂y ) (Persamaan ref {Ex6e5}) berkenaan dengan (y ):

[ start {align *} dfrac {∂ ^ 2f} {∂y ^ 2} & = dfrac {∂} {∂y} kiri [ dfrac {∂f} {∂y} kanan] angka [6pt] & = dfrac {∂} {∂y} [- 3xe ^ {- 3y} −5 cos (2x − 5y)] nonumber [6pt] & = 9xe ^ {- 3y} - 25 sin (2x − 5y). nonumber end {align *} nonumber ]

Latihan ( PageIndex {6} )

Hitung keempat-empat derivatif separa kedua untuk fungsi tersebut

[f (x, y) = sin (3x − 2y) + cos (x + 4y). nonumber ]

Petunjuk

Ikuti langkah yang sama seperti contoh sebelumnya.

Jawapan

( dfrac {∂ ^ 2f} {∂x ^ 2} = - 9 sin (3x − 2y) - cos (x + 4y) )

( dfrac {∂ ^ 2f} {∂x , ∂y} = 6 sin (3x − 2y) −4 cos (x + 4y) )

( dfrac {∂ ^ 2f} {∂y , ∂x} = 6 sin (3x − 2y) −4 cos (x + 4y) )

( dfrac {∂ ^ 2f} {∂y ^ 2} = - 4 sin (3x − 2y) −16 cos (x + 4y) )

Pada ketika ini kita harus perhatikan bahawa, di Contoh dan di pusat pemeriksaan, memang benar bahawa ( dfrac {∂ ^ 2f} {∂x∂y} = dfrac {∂ ^ 2f} {∂y∂x} ) . Dalam keadaan tertentu, ini selalu berlaku. Sebenarnya, ini adalah akibat langsung dari teorem berikut.

Kesamaan Derivatif Separa Campuran (Teorema Clairaut)

Katakan bahawa (f (x, y) ) didefinisikan pada cakera terbuka (D ) yang mengandungi titik ((a, b) ). Sekiranya fungsi (f_ {xy} ) dan (f_ {yx} ) berterusan pada (D ), maka (f_ {xy} = f_ {yx} ).

Teorema Clairaut menjamin bahawa selagi derivatif urutan kedua bercampur adalah berterusan, urutan di mana kita memilih untuk membezakan fungsi (iaitu, pemboleh ubah mana yang pertama, kemudian kedua, dan seterusnya) tidak penting. Ia juga boleh diperluas ke derivatif yang lebih tinggi. Bukti teorema Clairaut terdapat di kebanyakan buku kalkulus lanjutan.

Dua derivatif separa tertib kedua boleh dikira untuk sebarang fungsi (f (x, y). ) Derivatif separa (f_ {xx} ) sama dengan derivatif separa (f_x ) berkenaan dengan (x ), dan (f_ {yy} ) sama dengan terbitan separa (f_y ) berkenaan dengan (y ).

Lord Kelvin dan Zaman Bumi

Pada akhir 1800-an, para saintis bidang geologi baru sampai pada kesimpulan bahawa Bumi mesti berusia "berjuta-juta dan berjuta-juta" tahun. Pada waktu yang hampir sama, Charles Darwin telah menerbitkan risalah evolusi. Pandangan Darwin adalah bahawa evolusi memerlukan berjuta-juta tahun untuk berlaku, dan dia membuat tuntutan yang berani bahawa ladang kapur Weald, di mana fosil-fosil penting ditemui, adalah hasil hakisan (300 ) juta tahun.

Pada masa itu, ahli fizik terkemuka William Thomson (Lord Kelvin) menggunakan persamaan pembezaan separa penting, yang dikenali sebagai persamaan penyebaran haba, untuk menganggarkan usia Bumi dengan menentukan berapa lama masa yang diperlukan Bumi untuk menyejukkan dari batuan cair ke yang kita ada di masa itu. Kesimpulannya adalah pelbagai 20 ke 400 juta tahun, tetapi kemungkinan besar kira-kira 50 juta tahun. Selama beberapa dekad, pengisytiharan ikon sains yang tidak dapat disangkal ini tidak sesuai dengan ahli geologi atau dengan Darwin.

  • Baca kertas Kelvin mengenai menganggarkan usia Bumi.

Kelvin membuat andaian yang munasabah berdasarkan apa yang diketahui pada zamannya, tetapi dia juga membuat beberapa andaian yang ternyata salah. Satu anggapan yang salah adalah bahawa Bumi padat dan bahawa penyejukan oleh itu hanya melalui pengaliran, sehingga membenarkan penggunaan persamaan penyebaran. Tetapi kesalahan yang paling serius adalah kesalahan yang dapat dimaafkan - peninggalan fakta bahawa Bumi mengandungi unsur radioaktif yang sentiasa membekalkan haba di bawah mantel Bumi. Penemuan radioaktif hampir menjelang akhir kehidupan Kelvin dan dia mengakui bahawa pengiraannya harus diubah.

Kelvin menggunakan model satu dimensi sederhana yang hanya diterapkan pada cangkang luar Bumi, dan berasal dari grafik dan kecerunan suhu yang hampir diketahui di dekat permukaan Bumi. Mari kita lihat versi persamaan difusi yang lebih sesuai dalam koordinat radial, yang mempunyai bentuk

[ dfrac {∂T} {∂t} = K kiri [ dfrac {∂ ^ 2T} {∂ ^ 2r} + dfrac {2} {r} dfrac {∂T} {∂r} kanan ] label {kelvin1} ].

Di sini, (T (r, t) ) adalah suhu sebagai fungsi (r ) (diukur dari pusat Bumi) dan masa (t. K ) adalah kekonduksian haba — untuk batuan lebur, di kes ini. Kaedah standard untuk menyelesaikan persamaan pembezaan separa adalah dengan pemisahan pemboleh ubah, di mana kita menyatakan penyelesaiannya sebagai produk fungsi yang mengandungi setiap pemboleh ubah secara terpisah. Dalam kes ini, kita akan menuliskan suhu sebagai

[T (r, t) = R (r) f (t). ]

  1. Ganti borang ini menjadi Persamaan ref {kelvin1} dan, perhatikan bahawa (f (t) ) tetap sehubungan dengan jarak ((r) ) dan (R (r) ) adalah tetap berkaitan dengan masa ((t) ), tunjukkan bahawa [ dfrac {1} {f} dfrac {∂f} {∂t} = dfrac {K} {R} kiri [ dfrac {∂ ^ 2R} { ^r ^ 2} + dfrac {2} {r} dfrac {∂R} {∂r} kanan]. ]
  2. Persamaan ini mewakili pemisahan pemboleh ubah yang kita mahukan. Sisi kiri hanya fungsi (t ) dan sebelah kanan hanya fungsi (r ), dan mereka mesti sama untuk semua nilai (r ) dan (t ). Oleh itu, mereka berdua mesti sama dengan pemalar. Mari kita panggil pemalar (- λ ^ 2 ). (Kemudahan pilihan ini dilihat pada penggantian.) Oleh itu, kita mempunyai [ dfrac {1} {f} dfrac {∂f} {∂t} = - λ ^ 2 text {dan} dfrac {K } {R} kiri [ dfrac {∂ ^ 2R} {∂r ^ 2} + dfrac {2} {r} dfrac {∂R} {∂r} kanan] = - λ ^ 2. ]
  3. Sekarang, kita dapat mengesahkan melalui penggantian langsung untuk setiap persamaan bahawa penyelesaiannya adalah (f (t) = Ae ^ {- λ ^ 2t} ) dan (R (r) = B kiri ( dfrac { sin αr } {r} kanan) + C kiri ( dfrac { cos αr} {r} kanan) ), di mana (α = λ / sqrt {K} ). Perhatikan bahawa (f (t) = Ae ^ {+ λn ^ 2t} ) juga merupakan penyelesaian yang sah, jadi kami boleh memilih (+ λ ^ 2 ) untuk pemalar kami. Bolehkah anda melihat mengapa ia tidak berlaku untuk kes ini seiring bertambahnya waktu?
  4. Mari kita gunakan syarat sempadan.
    1. Suhu mesti terhad di tengah Bumi, (r = 0 ). Yang mana satu daripada dua pemalar, (B ) atau (C ), mestilah sifar untuk memastikan (R ) terhingga pada (r = 0 )? (Ingat bahawa ( sin (αr) / r → α = ) sebagai (r → 0 ), tetapi ( cos (αr) / r ) berkelakuan sangat berbeza.)
    2. Kelvin berpendapat bahawa ketika magma mencapai permukaan Bumi, ia menyejuk dengan cepat. Seseorang sering dapat menyentuh permukaan dalam beberapa minggu dari aliran. Oleh itu, permukaan mencapai suhu sederhana sangat awal dan tetap hampir tetap pada suhu permukaan (T_s ). Untuk kesederhanaan, mari kita tetapkan (T = 0 ) di (r = R_E ) dan cari α sehingga suhu ini ada sepanjang masa (t ). (Kelvin mengambil nilai menjadi (300K≈80 ° F ). Kami boleh menambahkan pemalar (300K ) ini ke penyelesaian kami nanti.) Untuk menjadi kenyataan ini, argumen sinus mestilah nol pada (r = R_E ). Perhatikan bahawa α mempunyai rangkaian nilai yang tidak terbatas yang memenuhi syarat ini. Setiap nilai (α ) mewakili penyelesaian yang sah (masing-masing dengan nilai tersendiri untuk (A )). Penyelesaian keseluruhan atau umum adalah jumlah semua penyelesaian ini.
    3. Pada (t = 0, ) kita menganggap bahawa seluruh Bumi berada pada suhu panas awal (T_0 ) (Kelvin menganggap ini sekitar (7000K ).) Penerapan keadaan sempadan ini melibatkan yang lebih maju penerapan pekali Fourier. Seperti yang dinyatakan di bahagian b. setiap nilai (α_n ) mewakili penyelesaian yang sah, dan penyelesaian umum adalah jumlah semua penyelesaian ini. Ini menghasilkan penyelesaian siri: [T (r, t) = kiri ( dfrac {T_0R_E} {π} kanan) sum_n dfrac {(- 1) ^ {n − 1}} {n} e ^ {- λn ^ 2t} dfrac { sin (α_nr)} {r}, teks {di mana} ; α_n = nπ / R_E ].

Perhatikan bagaimana nilai (α_n ) berasal dari keadaan sempadan yang diterapkan di bahagian b. Istilah ( dfrac {−1 ^ {n − 1}} {n} ) ialah pemalar (A_n ) untuk setiap istilah dalam siri ini, yang ditentukan daripada menggunakan kaedah Fourier. Membiarkan (β = dfrac {π} {R_E} ), memeriksa beberapa istilah pertama penyelesaian ini yang ditunjukkan di sini dan perhatikan bagaimana (λ ^ 2 ) dalam eksponen menyebabkan istilah yang lebih tinggi menurun dengan cepat seiring berjalannya waktu:

[T (r, t) = dfrac {T_0R_E} {πr} kiri (e ^ {- Kβ ^ 2t} ( sinβr) - dfrac {1} {2} e ^ {- 4Kβ ^ 2t} ( sin2βr) + dfrac {1} {3} e ^ {- 9Kβ ^ 2t} ( sin3βr) - dfrac {1} {4} e ^ {- 16Kβ ^ 2t} ( sin4βr) + dfrac {1 } {5} e ^ {- 25Kβ ^ 2t} ( sin5βr) ... kanan). ]

Masa hampir (t = 0, ) banyak istilah penyelesaian diperlukan untuk ketepatan. Memasukkan nilai untuk kekonduksian (K ) dan (β = π / R_E ) untuk masa yang menghampiri hanya ribuan tahun, hanya sebilangan istilah pertama yang memberikan sumbangan yang signifikan. Kelvin hanya perlu melihat penyelesaiannya di dekat permukaan Bumi (Gambar ( PageIndex {6} )) dan, setelah sekian lama, tentukan masa yang paling baik menghasilkan anggaran suhu yang diketahui pada zamannya ( (1 ° F ) ) kenaikan per (50 kaki )). Dia hanya memilih jangka masa dengan kecerunan yang hampir dengan nilai ini. Dalam Rajah ( PageIndex {6} ), penyelesaiannya diplot dan diskalakan, dengan suhu permukaan (300 − K ) ditambahkan. Perhatikan bahawa pusat Bumi agak sejuk. At the time, it was thought Earth must be solid.

Epilog

On May 20, 1904, physicist Ernest Rutherford spoke at the Royal Institution to announce a revised calculation that included the contribution of radioactivity as a source of Earth’s heat. In Rutherford’s own words:

“I came into the room, which was half-dark, and presently spotted Lord Kelvin in the audience, and realized that I was in for trouble at the last part of my speech dealing with the age of the Earth, where my views conflicted with his. To my relief, Kelvin fell fast asleep, but as I came to the important point, I saw the old bird sit up, open an eye and cock a baleful glance at me.

Then a sudden inspiration came, and I said Lord Kelvin had limited the age of the Earth, provided no new source [of heat] was discovered. That prophetic utterance referred to what we are now considering tonight, radium! Behold! The old boy beamed upon me.”

Rutherford calculated an age for Earth of about 500 million years. Today’s accepted value of Earth’s age is about 4.6 billion years.

Konsep kunci

  • A partial derivative is a derivative involving a function of more than one independent variable.
  • To calculate a partial derivative with respect to a given variable, treat all the other variables as constants and use the usual differentiation rules.
  • Higher-order partial derivatives can be calculated in the same way as higher-order derivatives.

Persamaan Utama

  • Partial derivative of (f) with respect to (x)

(dfrac{∂f}{∂x}=displaystyle{lim_{h→0}dfrac{f(x+h,y)−f(x,y)}{h}})

  • Partial derivative of (f) with respect to (y)

(dfrac{∂f}{∂y}=displaystyle{lim_{k→0}dfrac{f(x,y+k)−f(x,y)}{k}})

Glosari

higher-order partial derivatives
second-order or higher partial derivatives, regardless of whether they are mixed partial derivatives
mixed partial derivatives
second-order or higher partial derivatives, in which at least two of the differentiation are with respect to different variables
partial derivative
a derivative of a function of more than one independent variable in which all the variables but one are held constant

Determine the first partial derivatives of the function $f(x, y) = left<eginfrac<2x^3-y^3> & mathrm : (x, y) eq (0, 0)) 0 & mathrm (x, y) = (0, 0) end ight.$ , and determine if these partial derivatives are continuous at $(0, 0)$ .

Notice that in this example, $f$ is defined for all $(x, y) in mathbb^2$ , however, this function is a piecewise function, so we will deal with the partial derivatives of each piece of $f$ individually.

First suppose that $(x, y) eq (0, 0)$ . Then we can compute the partial derivatives of $f$ normally by applying the quotient rule:

Now suppose that $(x, y) = (0, 0)$ . We cannot use the partial derivative formulas above because we would get division by $ , so instead, we will use the formal definition of the partial derivative at $(0, 0)$ .

We will now determine whether or not these partial derivatives are continuous at $(0, 0) in mathbb^2$ . To show that, we must show whether or not $lim_ <(x, y) o (0, 0)>frac (x, y) = 2$ and whether or not $lim_ <(x, y) o (0, 0)>frac (x, y) = -frac<1><3>$ .

First let's determine if $frac$ is continuous at $(0, 0)$ .

Now notice that if we evaluate this limit along the line $x = 0$ then we get that $lim_ <(x, y) o (0, 0)>frac (x, y) = 0$ (which already tells us that our partial derivative is discontinuous), but also, if we evaluate this limit along the line $y = 0$ then we get that $lim_ <(x, y) o (0, 0)>frac (x, y) = 2$ . Since we have two different limits, we conclude that $lim_ <(x, y) o (0, 0)>frac (x, y)$ does not exist. Therefore we have that $frac$ is discontinuous at $(0, 0)$ .

Now let's determine if $frac$ is continuous at $(0, 0)$ .

Now notice that if we evaluate this limit along the line $x = 0$ we get that $lim_ <(x, y) o (0, 0)>frac (x, y) = - frac<1><3>$ . However, if we evaluate this limit along the line $y = 0$ we get that $lim_ <(x, y) o (0, 0)>frac (x, y) = 0$ . Since we have two different limits, we conclude that $lim_ <(x, y) o (0, 0)>frac (x, y)$ does not exist. Therefore we have that $frac$ is discontinuous at $(0, 0)$ .

Note that from Example 1 above, that a function can be differentiable even though its partial derivatives may be discontinuous.


Business Calculus with Excel

A standard technique in mathematics courses is to try to break a complicated problem into smaller and easier problems. For functions of several variables this can be done by looking at the variables one at a time, and treating the other variables as constants. Then we are back to considering functions of a single variable. We start by returning to Example 5 from the previous section, and seeing what information can be obtained by looking at one variable at a time.

Example 6.2.1 . Optimizing Revenue with Two Products.

I have a company that produces 2 products, widgets and gizmos. The two demand functions are:

This gives me the following revenue function:

Look at the functions of one variable obtained by treating either QG or QW as a constant. Use this information to find where we maximize revenue.

Solution: In terms of the last example, we want to start with a table and a wire frame chart.

The wires are obtained by intersecting the graph of the function with a plane where QW or QG is held constant.

Thus, when we treat either QW or QG as a constant we effectively are looking at one of the wires of the wire frame. To illustrate this, we will look at the wires corresponding to (QW=400) and (QG=300 ext<.>) When (QG=300 ext<,>) our revenue function simplifies to

Thus, the wire corresponding to (QG=300) is a parabola that bends down.

To find the vertex of the parabola, we take the derivative of our function of QW and set it equal to zero.

This derivative is zero when (QW=400 ext<.>) That is the only possible place on this wire where we can have a maximum.

This derivative is zero when (QG=250 ext<.>) That is the only possible place on this wire where we can have a maximum.

Putting the information together, the maximum must occur at (250,400). Putting these values back in the original equation gives a maximum of $5250 for the revenue function.

The procedure we used in the first example of replacing one variable with a constant and then taking the derivative of the resulting single variable function is a bit cumbersome. We can simplify the process by taking the derivative of the original function with respect to one variable while treating the other variables as constants. This is referred to as taking a partial derivative. There is also a change in notation. The familiar derivative of (f) with respect to (x) uses the symbol (frac f) while the partial derivative with respect to (x) uses the symbol (frac f ext<,>) or (f_x ext<.>) Similarly, the partial derivative with respect to (y) uses the symbol(frac f ext<,>) or (f_y ext<.>)

Example 6.2.2 . Finding and Interpreting Partial Derivatives.

Find the partial derivatives of (f(x,y)=x^2+ 2xy+3y^2-4x-3y) at ((x,y)=(3.5,-0.5) ext<.>) Explain what the partial derivatives mean in terms of the graph.

Solution: It is useful to look at a picture with the graph, the two curves obtained by keeping (x=3.5) and (y=1.5 ext<,>) and the tangent lines to those curves.

We also want to look at the slices corresponding keeping (x=-3.5) and (y=.5 ext<.>)

The yellow curve is obtained by fixing (y) and letting (x) vary. The blue curve is obtained by fixing (y) and letting (x) vary. We now take the partial derivatives with respect to both variables.

The partial derivatives give the slopes of the purple and red lines above. At the point ((3.5,-0.5) ext<,>) the (yellow) curves obtained by treating y as a constant and letting (x) vary has a (magenta) tangent line with a slope of (2 ext<,>) the value (frac f(3.5,-0.5) ext<.>) At the point ((3.5,-0.5) ext<,>) the (blue) curves obtained by treating (x) as a constant and letting (y) vary has a (red) tangent line with a slope of 1, the value (frac f(3.5,-0.5) ext<.>)

For functions of one variable, we had two main uses of the derivative. One was to identify candidate points for maxima and minima. We will look at critical points and extrema in the next section. The other use of the derivative was to produce a linear approximation or tangent line. We can generalize the tangent line for one variable to a tangent plane for two variables. For a function (f(x) ext<,>) we used the value of the point, ((x_0,f(x_0))) and the slope (f(x_0)) to get the equation of the tangent line approximation near (x_0 ext<.>)

For a function, (f(x,y) ext<,>) of two variables, we simply use partials for the slopes.

Example 6.2.3 . Approximating with a Tangent Plane.

The general Cobb-Douglas production function determines the Production (P), in terms of the variables Labor (L) and Capital (C):

or using short-hand notation:

where (c ext<,>) (alpha ext<,>) and (eta) are constants. For our widget factory, this becomes

with labor production and capital in the appropriate units. Find (Production(81,16) ext<.>) Use a linear approximation to estimate (Production(85,14) ext<.>)

Solution: We answer the first question by substituting the values into the equation.

To produce the tangent plane we take the partial derivatives and evaluate them at our base point.

This gives us our tangent plane.

Substituting in values gives our estimate.

In the case of the last example, evaluating the linear approximation was nicer than evaluating the function directly because the 4th roots of 16 and 81 are whole numbers, while the 4th roots of 85 and 14 are harder to compute. For real world functions, evaluating functions may involve a substantial investment of time and money, depending on the nature of the function.

In this section we have focused on functions of 2 variables since their graphs are surfaces in 3 dimensions, which is a familiar concept. For real world functions, we are often concerned with functions of many variables. The concept of partial derivative easily extends, with one variable and multiple parameters. Finding the linear approximation also extends without difficulty. We simply have a linear term for each variable.

Exercises Exercises: Wire Frames, Partial Derivatives, and Tangent Planes Problems

For exercises 1-7, for the given functions and points (P_1) and (P_2 ext<:>)

Give the 2 functions of one variable through (P_1) obtained by holding each variable constant.

Find the partial derivatives of the original function.

Evaluate the partial derivatives at (P_1 ext<.>)

Give the equation of the tangent plane through (P_1 ext<.>)

The approximation at (P_2) obtained from the tangent plane.

The function is (f(x,y)=x^2+3xy+4y^2 ext<,>) (P_1=(4,2) ext<,>) and (P_2=(3,2.5) ext<.>)

Give the 2 functions of one variable through (P_1) obtained by holding each variable constant.

Find the partial derivatives of the original function.

Evaluate the partial derivatives at (P_1 ext<.>)

Give the equation of the tangent plane through (P_1 ext<.>)

We need (f(4,2)=16+24+16=56) for the equation of the tangent plane.

The approximation at (P_2) obtained from the tangent plane.

The function is (f(x,y)=(x+3y)/(x^2+y^2 ) ext<,>) (P_1=(2,3) ext<,>) and (P_2=(3,2.5) ext<.>)

The function is (f(x,y)=(x^2)(x+2^y) ext<,>) (P_1=(3,-1) ext<,>) and (P_2=(3,0) ext<.>)

Give the 2 functions of one variable through (P_1) obtained by holding each variable constant.

Find the partial derivatives of the original function.

Evaluate the partial derivatives at (P_1 ext<.>)

Give the equation of the tangent plane through (P_1 ext<.>)

We need (f(3,-1)=9*frac<7><2>=frac<63><2>) for the equation of the tangent plane

The approximation at (P_2) obtained from the tangent plane.

The function is the revenue function for selling widgets and gizmos with demand price functions

and, (P_1=(QuanityGizmos,QuantityWidgets)=(1000,500) ext<,>) and (P_2=(1050,575) ext<.>)

The function is the revenue function for selling widgets and gizmos with demand price functions

and, (P_1=(QuanityGizmos,QuantityWidgets)=(800,400) ext<,>) and (P_2=(750,425) ext<.>)

For the sake of notation we will use the following abbreviations:

and, (P_1=(QG,QW)=(800,400) ext<,>) and (P_2=(750,425) ext<.>)

We need to find the Revenue function to solve the problem:

Give the 2 functions of one variable through (P_1) obtained by holding each variable constant.

This function gives us information about the revenue in terns of Widgets near a production level of 400 widgets and 800 gizmos. We can use Wolfram Alpha to graph this. Assuming there are 800 gizmos the widget influence on the revenue looks like this

The slope is about (m = 4 ext<.>)

The revenue generated by the gizmos assuming the number of widgets = 400 and the number of gizmos is near 800 gives the following picture

The slope is about (m=8 ext<.>)

Find the partial derivatives of the original function.

In part a we saw that the revenue function seems to be growing faster for the gizmo variable, then for the widget variable. To get more information we can compute the partial derivatives (b) and then evaluate them at (P_1) (c).

Note that the first part requires a product rule and then a chain rule to deal with the exponential part of the formula.

Evaluate the partial derivatives at (P_1 ext<.>)

The estimates we observed in part a were fairly close to the actual rates of change.

Give the equation of the tangent plane through (P_1 ext<.>)

We need to find (Rev(800,400) ext<.>) Using Wolfram Alpha (or calculator) we get

The approximation at (P_2) obtained from the tangent plane.

The estimated revenue when (P_2=(750,425)) is given by

In this case the change in production would result in a loss in revenue. This is mainly due to the impact of the lower production in gizmos.

The function is the Cobb-Douglas production function in a widget factory,

where labor is in workers, capital equipment is in units of $20,000, and production is in units of 200 widgets produced per month. In the ((Labor,Capital)) plane, let (P_1=(100,30) ext<,>) and (P_2=(110,25) ext<.>)

The function is the Cobb-Douglas production function in a country,

where labor is in millions of workers, capital equipment is in units of billions of dollars, and production is in units of billions of dollars per year. In the ((Labor,Capital)) plane, let (P_1=(300,30) ext<,>) and (P_2=(310,32) ext<.>)

Give the 2 functions of one variable through (P_1) obtained by holding each variable constant.


6.3: Partial Derivatives - Mathematics

The Maple commands for computing partial derivatives are D and diff . The diff command can be used on both expressions and functions whereas the D command can be used only on functions. The commands below show examples of first order and second order partials in Maple.

Note in the above D command that the 1 in the square brackets means x and the 2 means y . The next example shows how to evaluate the mixed partial derivative of the function given above at the point .

To find a point, , where the tangent plane is horizontal, you would need to solve where both first order partials are equal to zero simultaneously.

The horizontal plane at that point would simply be . Below is how to plot the surface and the horizontal tangent plane.

To find the tangent plane to the sphere at the point and is positive, you would first need to find the coordinating value for the ordered pair. Below is how you would do this in Maple as well as find and plot the tangent plane implicitly.

at the point using the diff command and then again using the D command.

a) Find the plane tangent to the given surface at . b) Plot the surface and the tangent plane on the same graph and rotate the 3-D plot to show the point of tangency. Use plotting ranges and . c) Find the point where the tangent plane to the given surface would be horizontal. d) Plot the surface and the horizontal tangent plane on the same graph and rotate the 3-D plot to show the point of tangency. Use the same plotting ranges as above.


Bezier triangle partial derivatives

Hello I am trying to compute the partial derivatives of a bezier triangle (in u and v directions).

I followed this article to build my program.

The section that interest me is " Derivatives [triangles]:"

question 1 : In the equations. does "p" refer to the actual control points of the triangle?

If this is the case, it seems like the derivative takes into account only 6 of the 10 control points of the patch. Is this normal?

the coefficients computed are for i+j+k=2 so if I plug theses values in the bernstein polynomial definition I get the following coefficients :


i=0, j=0, k=2 ---> w * w
i=0, j=2, k=0 ---> v * v
i=2, j=0, k=0 ---> u * u
i=1, j=1, k=0 ---> 2 * u * v
i=1, j=0, k=1 ---> 2 * u * w
i=0, j=1, k=1 ---> 2 * v * w

question 2 : this seems to contradict another section of the article where a bernstein of degree two is used for indices 110, 101, 011 and there is no x2 multiplication. Is this normal?

Hopefully someone can clarify this. I really need theses derivatives.

According to the terminology used so far, I’d say yes.

I found an old paper by Sederberg, which has a graphical explanation for the definition of directional derivatives (Fig. 14). The figure depicts the construction of the derivative patch in w-direction (w=const). The construction for the patches in u- and v-direction is similar. Though, it has a little error in there, I think. The top and top left control points are substituted, but I think you can get the gist.

It’s expected that the degree of the “derivative patch” goes down, so 6 points instead of 10 seems right. If you derive a polynomial of degree n the outcome is a polynomial of degree n-1.

This is indeed a little strange. I’d have expected the factor 2, too.
Well, after all, this article seems not to be peer-reviewed, so little errors might be possible. I’d suggest to pick up a book of Farin, e.g. “Curves and Surfaces for CAGD: A Practical Guide” (Chapter 17) if you want to learn more.

I manually derived the position equation to find the partial derivatives.


For n=3 the code looks like this :


float3 bezierTriangle(float3 uvw, float3 points[10])
<
return
uvw.x * uvw.x * uvw.x * points[P300] +
uvw.y * uvw.y * uvw.y * points[P030] +
uvw.z * uvw.z * uvw.z * points[P003] +
3.0f * uvw.x * uvw.x * uvw.y * points[P210] +
3.0f * uvw.y * uvw.y * uvw.z * points[P021] +
3.0f * uvw.z * uvw.z * uvw.x * points[P102] +
3.0f * uvw.y * uvw.y * uvw.x * points[P120] +
3.0f * uvw.z * uvw.z * uvw.y * points[P012] +
3.0f * uvw.x * uvw.x * uvw.z * points[P201] +
6.0f * uvw.x * uvw.y * uvw.z * points[P111]
>

. And the derivatives in u and v are :


float3 bezierTriangle_du(float3 uvw, float3 points[10])
<
return
3.0f * uvw.x * uvw.x * (points[P300] - points[P201]) +
3.0f * uvw.y * uvw.y * (points[P120] - points[P021]) +
3.0f * uvw.z * uvw.z * (points[P102] - points[P003]) +
6.0f * uvw.x * uvw.y * (points[P210] - points[P111]) +
6.0f * uvw.x * uvw.z * (points[P201] - points[P102]) +
6.0f * uvw.y * uvw.z * (points[P111] - points[P012])
>

float3 bezierTriangle_dv(float3 uvw, float3 points[10])
<
return
3.0f * uvw.x * uvw.x * (points[P210] - points[P201]) +
3.0f * uvw.y * uvw.y * (points[P030] - points[P021]) +
3.0f * uvw.z * uvw.z * (points[P012] - points[P003]) +
6.0f * uvw.x * uvw.y * (points[P120] - points[P111]) +
6.0f * uvw.x * uvw.z * (points[P111] - points[P102]) +
6.0f * uvw.y * uvw.z * (points[P021] - points[P012])
>

As we can see, the derivatives for a degree 3 Bezier triangle is a degree 2 Bezier triangle. But the control points are not like they describe in the article.

We actually use 9 of the 10 initial control points to make 6 new ones.

I guess it could be written like this :

I also compared the results of adding a a factor 2 to the original N-Patch normal equation. Obviously, it still works without it but I think I looks better with the factor 2.


Partial Derivative Calculator


Partial Derivative Calculator computes derivatives of a function with respect to given variable utilizing analytical differentiation and displays a step-by-step solution. It gives chance to draw graphs of the function and its derivatives. Calculator maintenance derivatives up to 10th order, as well as complex functions. Derivatives being computed by parsing the function, utilizing differentiation rules and simplifying the result.

How to Use Partial Derivative Calculator

1. Enter every needed function to differentiate
2. Enter differentiation variable if it is different from the the default value
3. Choose degree of differentiation
4. Click Compute button
5. Derivative of the function will be computed and displayed on the screen
6. Click on Show a step by step solution if there is a desire to observe the differentiation steps
7. Click on Draw graph to display graphs of the function and its derivative

This calculator can take the partial derivative of regular functions, as well as trigonometric functions.


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    My research showed only that it's called "the curved d" or sometimes "the curly d."

    It is not the lower case delta (the Greek letter).

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      Pilihan

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      Look at the complete list of languages: Available language pairs

      There are two Japanese-English (and Japanese-French) dictionaries and one contains Kanji and Kana (Kana in English and French pair due to improved searching). For the same reason the Chinese dictionary contains traditional and simplified Chinese terms on one side and Pinyin and English terms on the other.

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      Partial Derivatives of a unit vector

      to provide more background, I'm trying to find the derivative of the constraint function:

      C(p1,p2,p3) = arccos( PerpDot(n1,n2) ) - theta

      where n1 = ((p2 - p1)/||p2 - p1||) and n2 = ((p3 - p1)/||p3 - p1||)

      the theta term disappears in the derivative, and the derivative of arccos(x) is -1/(sqrt(1 - x^2)). So, all that remains is to figure out the derivative of the PerpDot.

      If i understand correctly, perpdot is really just the inner product in 2D:
      perpdot(a,b) = dot(a^T,b)

      The function is a constraint function which describes the error between a current angle and some desired angle. I'm trying to understand this paper: http://graphics.ethz.ch/

      Let p2 = (x,y) and p1 = (a,b), where x and y are the variables and where a and b are constants. Then n = (y-b,-(x-a))/pow((x-a)^2+(y-b)^2,1/2).

      The derivative with respect to x is the vector

      and the derivative with respect to y is the vector

      The first derivative of n with respect to (x,y) is written as a matrix whose columns are dn/dx and dn/dy.

      Is my simplification (translating everything to locate p1 at the origin) valid? Also, do you know of a good book/site for learning this sort of thing? I've found many engineering tutorials, but they never follow through all the way to show how the derivatives are actually computed!

      In the paper I mentioned they derive the gradient of the normalized cross product n = (p1 x p2)/||p1 x p2||

      I'm trying to do the same thing, but in 2D, so I need the gradient of the normalized perpdot am I on the right track? I feel a bit stupid.. this math seems like it makes perfect sense, but it's still over my head I'm afraid!

      Basically in a partial derivative you're just treating all the variables except for one as constants. For example, the partial derivative of f(x,y,z) = xyz with respect to x is just yz.

      Just expand out this perpdot thing by substitution using its definition and it might become a lot clearer. perpdot(a,b) = a.y * b.x - a.x * b.y, I think. So if b is a constant, the derivative with respect to a.x is -b.y.

      I'm a bit confused about vector-valued functions -- it seems like i should just treat each vector as two separate variables, but I'm not sure that this is correct.

      In the paper I linked to they're using the gradient of a function wrt a _vector_, not a component of a vector: is this just shorthand for finding the gradient wrt each of the vector's components?

      thanks so much for the info.

      You must distiguish between scalar and vector valued functions. E.g.

      C1(p1,p2) = p1 * p2 is a scalar valued function
      C2(p1,p2) = p1 x p2 is a vector valued function

      Since p1 and p2 are functions of the time you can find the Jacobians by inspection

      dC2/dt = p1 x v2 + v1 x p2 = p1 x v2 - p2 x v1 = P1 * v2 - P2 * v1 where P_i is the cross-matrix of p_i

      You see that the "gradient" for the vector valued functions is a vector. While the so called "vector gradient" is a matrix (tensor) often refered to as Jacobian.

      Regarding your specific constraint funtions. If it is from the "Position Based Dynamics" paper there is a derivation in the appendix of the paper.

      I also found it very difficult to find material about this topic and I can recommend only a few things

      a) The math reference of E. Guendelman was very helpful
      http://graphics.stanford.edu/

      b) This paper of R. Bridson has some information in the appendix
      http://www.cs.ubc.ca/

      Scalar fields
      Vector fields
      Gradient
      Jacobian

      This should get you a little further. Mostly it behaves like normal analysis. The tricky part is the multiplication order and you often have to use the transpose. See section "2 Multivariable" in the Guendelman math reference.


      Partial Differentiation Questions and Answers – Variable Treated as Constant

      This set of Differential and Integral Calculus Multiple Choice Questions & Answers (MCQs) focuses on “Variable Treated as Constant”.

      1. If z=3xy+4x 2 , what is the value of (frac<∂z><∂x>)?
      a) 3y+8x
      b) 3x+4x 2
      c) 3xy+8x
      d) 3y+3x+8x
      View Answer

      2. The value of (frac<∂z><∂y>)=8x 2 +6xy 2 +4. What is the function z expressed as?
      a) z=8x 3 +2x 2 y 2 +4x
      b) z=8x 2 y+2xy 3 +4y
      c) z=8y+2xy 2 +4y
      d) z=16x+6y 2
      View Answer

      4. Volume of an object expressed in spherical coordinates is given by (V = int_0^<2>>∫_0^<3>>∫_0^1r cos∅ dr d∅ dθ.) The value of the integral is _______
      a) (frac><2>)
      b) (frac<1> <√2>π)
      c) (frac> <2>π)
      d) (frac><8>π)
      View Answer

      6. Berapakah nilai ( frac <∂ ^ 2z> <∂x∂y> ) untuk z = 3x 2 y + 5y?
      a) 3xy
      b) 6x
      c) 3x + 5
      d) 6xy
      Lihat Jawapan

      8. Diberi (∫_0 ^ 8x ^ < frac <1> <3>> dx, ) cari ralat dalam menghampiri kamiran menggunakan Peraturan 1/3 Simpson dengan n = 4.
      a) 1.8
      b) 2.9
      c) 0.3
      d) 0.35
      Lihat Jawapan

      9. Penentu matriks yang nilai eigennya adalah 6, 4, 3 diberikan oleh ___________
      a) 3
      b) 24
      c) 72
      d) 13
      Lihat Jawapan

      10. Simbol yang digunakan untuk derivatif separa, ∂, pertama kali digunakan dalam matematik oleh Marquis de Condorcet.
      a) Betul
      b) Salah
      Lihat Jawapan

      Sanfoundry Global Education & # 038 Seri Pembelajaran - Perbezaan dan Kalkulus Integral.

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