BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.4, Problem 62E
To determine

To find: The derivative f(x) in terms of g,g and g.

Expert Solution

Answer to Problem 62E

The derivative f(x) in terms of g,g and g is f(x)=4x3g(x2)+6xg(x2).

Explanation of Solution

Given:

The function is f(x)=xg(x2).

Result used: 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:

Product Rule:ddx[f(x)g(x)]=f(x)ddx[g(x)]+g(x)ddx[f(x)]

Sum Rule:ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)]

Calculation:

Obtain the first derivative of f(x)=xg(x2).

f(x)=ddx(f(x))=ddx(xg(x2))

Apply the product rule (1),

f(x)=xddx(g(x2))+g(x2)ddx(x)=xddx(g(x2))+g(x2)(1x11)

=xddx(g(x2))+g(x2) (2)

Obtain the derivative ddx(g(x2)) by using the chain rule as shown in equation (1).

Let h(x)=x2 and f(u)=g(u)  where u=h(x).

ddx(g(x2))=g(h(x))h(x) (3)

The derivative of f(h(x)) is computed as follows,

f(h(x))=f(u)=ddu(f(u))=ddu(g(u))=g(u)

Substitute u=x2 in the above equation,

f(h(x))=g(x2)

Thus, the derivative is f(h(x))=g(x2).

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

h(x)=ddx(x2)=2x21=2x

Thus, the derivative is h(x)=2x.

Substitute g(x2) for f(h(x)) and 2x for h(x) in equation (3),

ddx(g(x2))=g(x2)2x=2xg(x2)

Substitute 2xg(x2) for ddx(g(x2)) in equation (2),

  f(x)=x(2xg(x2))+g(x2)=2x2g(x2)+g(x2)

Therefore, the derivative of f(x)=xg(x2) is f(x)=2x2g(x2)+g(x2).

Obtain the second derivative of f(x).

f(x)=ddx(f(x))=ddx(2x2g(x2)+g(x2))

Apply the sum rule (2),

f(x)=ddx(2x2g(x2)+g(x2))

=ddx(2x2g(x2))+ddx(g(x2)) (4)

Obtain the derivative ddx(2x2g(x2)) by using the product rule (1),

ddx(2x2g(x2))=2x2ddx[g(x2)]+g(x2)ddx[2x2]=2x2ddx[g(x2)]+g(x2)2[2x21]

                   =2x2ddx[g(x2)]+g(x2)[4x] (5)

Obtain the derivative ddx(g(x2)) by using the chain rule as shown in equation (1).

Let h(x)=x2 and k(u)=g(u)  where u=h(x).

ddx(g(x2))=k(h(x))h(x) (6)

The derivative of k(h(x)) is computed as follows,

k(h(x))=k(u)=ddu(k(u))=ddu(g(u))=g(u)

Substitute u=x2 in the above equation,

g(h(x))=g(x2)

Thus the derivative is g(h(x))=g(x2).

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

h(x)=ddx(x2)=2x21=2x

Thus the derivative is h(x)=2x.

Substitute g(x2) for k(h(x)) and 2x for h(x) in equation (6),

ddx(g(x2))=g(x2)2x=2xg(x2)

Substitute 2xg(x2) for ddx(g(x2)) in equation (5),

ddx(2x2g(x2))=2x2(2xg(x2))+g(x2)[4x]=4x3g(x2)+4xg(x2)

Substitute 4x3g(x2)+4xg(x2) for ddx(2x2g(x2)) and 2xg(x2) for ddx(g(x2)) in equation (4),

f(x)=4x3g(x2)+4xg(x2)+2xg(x2)=4x3g(x2)+6xg(x2)

Therefore, the derivative f(x) in terms of g,g and g is

f(x)=4x3g(x2)+6xg(x2).

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