   Chapter 14, Problem 24RE

Chapter
Section
Textbook Problem

If z = sin(x + sin t), show that ∂ z ∂ x ∂ 2 z ∂ x ∂ t = ∂ z ∂ t ∂ 2 z ∂ x 2

To determine

To show: The equation zx2zxt=zt2zx2 if z=sin(x+sint) .

Explanation

Given:

The function is z=sin(x+sint) .

Calculation:

The function is z=sin(x+sint) (1)

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

zx=x(sin(x+sint))=cos(x+sint)(1+0)=cos(x+sint)

Thus, the value of zx is cos(x+sint) .

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

zt=t(sin(x+sint))=cos(x+sint)(0+cost)=cos(x+sint)(cost)

Thus, the value of zt is cos(x+sint)(cost) .

Take partial derivative with respect to x in the equation zt=cos(x+sint)(cost) ,

2zxt=x[cos(x+sint)(cost)]=[sin(x+sint)(cost)(1+0)]=sin(x+sint)(cost)

Thus, the value of 2zxt is sin(x+sint)(cost) .

Take partial derivative with respect to x in the equation zx=cos(x+sint) ,

2zx2=x[cos(x+sint)]=[sin(x+sint)(1+0)]=sin(x+sint)

Thus, the value of 2zx2 is sin(x+sint)

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