   Chapter 14, Problem 42RE

Chapter
Section
Textbook Problem

If cos(xyz) = 1 + .x2y2 + z2, find ∂ z ∂ x and ∂ z ∂ y .

To determine

To find: The values of zxandzy if cos(xyz)=1+x2y2+z2 .

Explanation

Given:

The function is, cos(xyz)=1+x2y2+z2 .

Equation 7:zx=FxFz=FxFzandzy=FyFz=FyFz , where F is the function of x,yandz ”.

Calculation:

Let the function be, F(x,y,z)=1+x2y2+z2cos(xyz) (1)

The equations of zxandzy using the equation 7 as follows,

zx=FxFz (2)

zy=FyFz (3)

Take partial derivative with respect to x in the equation (1),

Fx=x(1+x2y2+z2cos(xyz))Fx=0+2xy2+0+sin(xyz)(yz)=2xy2+yzsin(xyz)

Thus, the partial derivate of x , Fx=2xy2+yzsin(xyz) .

Take the partial derivative with respect to y in the equation (1),

Fy=y(1+x2y2+z2cos(xyz))Fy=0+2x2y+0(sin(xyz)(xz))=2x2y+xzsin(xyz)

Thus, the partial derivate of y , Fy=2x2y+xzsin(xyz)

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