   Chapter 14.5, Problem 21E

Chapter
Section
Textbook Problem

Use the Chain Rule to find the indicated partial derivatives.21. z = x4 + x2y, x = s + 2t – u, y = stu2; ∂ z ∂ s , ∂ z ∂ t , ∂ z ∂ u when s = 4, t = 2, u = 1

To determine

To find: The value of zs,zt and zu using the chain rule if z=x4+x2y,x=s+2tu,y=stu2 when s=4,t=2andu=1 .

Explanation

The function is, z=x4+x2y .

Given that the value of s=4,t=2, and u=1 .

Substitute s=4,t=2, and u=1 in x,

x=s+2tu=4+2(2)1=4+41=7

Thus, the value of x=7 .

Substitute s=4,t=2, and u=1 in y,

y=stu2=4(2)(12)=8

Thus, the value of y=8 .

The partial derivative zs using chain rule is computed as follows,

zs=zxxs+zyys=x(x4+x2y)s(s+2tu)+y(x4+x2y)s(stu2)=(4x3+2xy)(1)+(x2)(tu2)=4x3+2xy+tx2u2

Thus, the partial derivative, zs=4x3+2xy+tx2u2 .

Substitute the respective values  in zs and obtain the required value.

zs=4(7)3+2(7)8+(2)(7)2(1)2=1372+112+98=1582

Thus, the required value is, zs=1582

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