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Mathematics 1
Mathematics 1 Deneme Sınavı
Mathematics 1 Deneme Sınavı Sorusu #1233100
Mathematics 1 Deneme Sınavı Sorusu #1233100
Given that z=(x+y)2, y=u3 x=u2-u find the partial derivative dz/du
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9u5-12u4+5u3+6u2-2u |
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9u5+12u4-5u3-6u2+2u |
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6u5-5u3-6u2+2u |
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6u5+5u4-5u3-6u2+2u |
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u5+12u4-u3-6u2+2 |
Yanıt Açıklaması:
dz/du=(dz/dx)*(dx/du)+(dz/dy)*(dy/du)
Since z=(x+y)2, dz/dx=2x+2y and dz/dy=2x+2y
Since y=u3, dy/du=3u2
x=u2-u, dx/du=2u-1
By substitution
dz/du=(2x+2y)(2u-1)+(2x+2y)3u2=(2x+3y)(3u2+2u-1)=(2(u2-u)+3u3)(3u2+2u-1)=(2u2-2u+3u3)(3u2+2u-1)=9u5+12u4-5u3-6u2+2u
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