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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 3.3, Problem 30E

(a)

To determine

To find: The differentiation of g(x)=f(x)sinx.

Expert Solution

Answer to Problem 30E

The differentiation of g(x)=f(x)sinx at x=π3 is 23.

Explanation of Solution

Given:

The function is g(x)=f(x)sinx and f(π3)=4, f(π3)=2

Formula used:

Product Rule:

Product rule for two functions f1(x) and f2(x) is given as follows.

ddx[g1(x)g2(x)]=g1(x)ddx[g2(x)]+g2(x)ddx[g1(x)] (1)

Calculation:

Apply Product Rule as shown in equation (1).

In equation (1), substitute f(x) for g1(x) and sinx for g2(x).

ddx(f(x)sinx)=f(x)ddx(sinx)+sinxddx(f(x))=f(x)(cosx)+sinx(f(x))   

ddx(f(x)sinx)=f(x)cosx+f(x)sinx    (2)

In equation (2), substitute x=π3.

g(π3)=f(π3)cosπ3+f(π3)sinπ3

Substitute 4 for f(π3) and –2 for f(π3).

g(π3)=4cosπ32sinπ3=4(12)+(2)(32)      (cosπ3=12,sinπ3=32)=23   

Therefore, The differentiation of g(x)=f(x)sinx at x=π3 is 23.

(b)

To determine

To find: The differentiation of h(x)=cosxf(x).

Expert Solution

Answer to Problem 30E

The differentiation of h(x)=cosxf(x) at x=π3 is 23+116_

Explanation of Solution

Given:

The function is h(x)=cosxf(x) and f(π3)=4, f(π3)=2.

Formula used:

Quotient rule:

If f(x) and g(x) are differentiable function, then the quotient rule is,

ddx[f(x)g(x)]=g(x)ddx[f(x)]f(x)ddx[g(x)][g(x)]2 (3)

Calculation:

Apply Quotient Rule as shown in equation (3).

Substitute cosx for f(x) and f(x) for g(x) in equation (3).

ddx[cosxf(x)]=[f(x)]ddx[cosx]cosxddx[f(x)]     [f(x)]2 =[f(x)][sinx]cosx[f(x)]     [f(x)]2=f(x)sinxf(x)cosx     [f(x)]2

In equation (3), substitute x=π3

h(π3)=f(π3)sinπ3f(π3)cosπ3  [f(π3)]2

Substitute 4 for f(π3) and –2 for f(π3).

h(π3)=4(32)(2)12[4]2          ( cosπ3=12,sinπ3=32)=23+1[4]2=23+116

Therefore, the differentiation of h(x)=cosxf(x) at x=π3 is 23+116_.

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