   Chapter 3.4, Problem 21E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find the derivative of the function. y = x x + 1

To determine

To find: The derivative of the function y=xx+1.

Explanation

Given:

The function is y=xx+1.

Result used:

The Chain Rule:

If h is differentiable at x and g is differentiable at h(x), then the composite function F=gh defined by F(x)=g(h(x)) is differentiable at x and F is given by the product

F(x)=g(h(x))h(x) (1)

Derivative Rules:

(1) Power Rule: ddx(xn)=nxn1.

(2) Quotient Rule: ddx[f(x)g(x)]=g(x)ddx[f(x)]f(x)ddx[g(x)][g(x)]2.

Calculation:

Obtain the derivative of y.

y=ddx(y)=ddx(xx+1)=ddx((xx+1)12)

Let h(x)=xx+1 and g(u)=u  where u=h(x).

Apply the chain rule as shown in equation (1),

y=g(h(x))h(x) (2)

The derivative g(h(x)) is computed as follows,

g(h(x))=g(u)         [Qu=h(x)]=ddu(g(u))=ddu(u)

Apply the power rule (1),

g(h(x))=12u121=12u12=12u12=12u

Substitute u=xx+1 and simplify further

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