z=x+2y+f(3x-4y),z ▏y=0时=x^2,求偏导Z/偏导x和z对y的偏导
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z=x+2y+f(3x-4y),z ▏y=0时=x^2,求偏导Z/偏导x和z对y的偏导
2/3(3x-4y) 和 10/3-8/9(3x-4y)
2/3(3x-4y) 和 10/3-8/9(3x-4y)
y=0
z = x+2*0+f(3x-0) = x^2
=> x+f(3x)=x^2
=> f(3x)=x^2-x
即 f(t)= t^2/9-t/3
f'(t)= 2t/9-1/3
z'x = x'+2y'+f'(3x-4y)
= 1+0+(3x-4y)'*(2*(3x-4y)/9-1/3)
= 1+3*((2x/3-8y/9)-1/3)
= 1+2x-8y/3-1
= 2x-8y/3
=2/3 * (3x-4y)
z'y = 0+2 +(3x-4y)'*(2(3x-4y)/9-1/3)
= 2-4*(2(3x-4y)/9-1/3)
= 2- 8/9*(3x-4y) -4/3
= 10/3 - 8/9(3x-4y)
z = x+2*0+f(3x-0) = x^2
=> x+f(3x)=x^2
=> f(3x)=x^2-x
即 f(t)= t^2/9-t/3
f'(t)= 2t/9-1/3
z'x = x'+2y'+f'(3x-4y)
= 1+0+(3x-4y)'*(2*(3x-4y)/9-1/3)
= 1+3*((2x/3-8y/9)-1/3)
= 1+2x-8y/3-1
= 2x-8y/3
=2/3 * (3x-4y)
z'y = 0+2 +(3x-4y)'*(2(3x-4y)/9-1/3)
= 2-4*(2(3x-4y)/9-1/3)
= 2- 8/9*(3x-4y) -4/3
= 10/3 - 8/9(3x-4y)
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