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 3.2, Problem 48E

(a)

To determine

To find: The expression for the derivative of the function y=x2f(x).

Expert Solution

Answer to Problem 48E

The expression for the derivative of the function y=x2f(x) is dydx=x2f(x)+2xf(x).

Explanation of Solution

Given:

The function is y=x2f(x).

Derivative rule:

(1) Product Rule: ddx[f1(x)f2(x)]=f1(x)ddx[f2(x)]+f2(x)ddx[f1(x)]

(2) Quotient Rule: If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]f1(x)ddx[f2(x)][f2x]2

(3) Power rule: ddx(xn)=nxn1

(4) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

Calculation:

Obtain the derivative of the function y=x2f(x) as follows,

dydx=ddx(y)=ddx(x2f(x))

Apply the product rule (1) and the power rule (3) as,

dydx=x2ddx(f(x))+f(x)ddx(x2)=x2f(x)+f(x)(2x)=x2f(x)+2xf(x)

Therefore, the expression for the derivative of the function y=x2f(x) is dydx=x2f(x)+2xf(x).

(b)

To determine

To find: The expression for the derivative of the function y=f(x)x2.

Expert Solution

Answer to Problem 48E

The expression for the derivative of the function y=f(x)x2 is dydx=xf(x)2f(x)x3.

Explanation of Solution

Given:

The function is y=f(x)x2.

Calculation:

Obtain the derivative of the function y=f(x)x2 as follows,

dydx=ddx(f(x)x2)

Apply the quotient rule (2) and the power rule (3) as,

dydx=x2ddx(f(x))f(x)ddx(x2)(x2)2=x2(f(x))f(x)(2x)x4=x(xf(x)2f(x))x4=xf(x)2f(x)x3

Therefore, the expression for the derivative of the function y=f(x)x2 is dydx=xf(x)2f(x)x3.

(c)

To determine

To find: The expression for the derivative of the function y=x2f(x).

Expert Solution

Answer to Problem 48E

The expression for the derivative of the function y=x2f(x) is dydx=2xf(x)x2f(x)(f(x))2.

Explanation of Solution

Given:

The function is y=x2f(x).

Calculation:

Obtain the derivative of the function y=x2f(x) as follows,

dydx=ddx(y)=ddx(x2f(x))

Apply the quotient rule (2) and the power rule (3) as,

dydx=f(x)ddx(x2)x2ddx(f(x))(f(x))2=f(x)(2x)x2(f(x))(f(x))2=2xf(x)x2f(x)(f(x))2

Therefore, the expression for the derivative of the function y=x2f(x) is dydx=2xf(x)x2f(x)(f(x))2.

(d)

To determine

To find: The expression for the derivative of the function y=1+xf(x)x.

Expert Solution

Answer to Problem 48E

The expression for the derivative of the function y=1+xf(x)x is dydx=2x2f(x)+xf(x)12x32.

Explanation of Solution

Given:

The function is y=1+xf(x)x.

Calculation:

The given function can be expressed as follows,

y=1+xf(x)x12=1x12+xf(x)x12=x12+x112f(x)=x12+x12f(x)

Obtain the derivative of the function y=1+xf(x)x as follows,

dydx=ddx(y)=ddx(x12+x12f(x))

Apply the sum rule (4) and the product rule (1) as,

dydx=ddx(x12)+ddx(x12f(x))=ddx(x12)+[x12ddx(f(x))+f(x)ddx(x12)]

Apply the power rule (3) and simplify the terms,

dydx=(12x121)+[x12(f(x))+f(x)(12x121)]=12x32+x12f(x)+12f(x)x12=12x32+2x2f(x)2x2x12+xf(x)2xx12=12x32+2x2f(x)2x32+xf(x)2x32

Simplify the terms and obtain the derivative of the function,

dydx=1+2x2f(x)+xf(x)2x32=2x2f(x)+xf(x)12x32

Therefore, the expression for the derivative of the function y=1+xf(x)x is dydx=2x2f(x)+xf(x)12x32.

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