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 3.3, Problem 29E

(a)

To determine

To find: The derivative of f(x)=tanx1secx.

Expert Solution

Answer to Problem 29E

The derivative of f(x) is 1+tanxsecx_.

Explanation of Solution

Given:

The function is f(x)=tanx1secx.

Derivative rule: Quotient rule

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

ddx[f(x)g(x)]=g(x)ddx[f(x)]f(x)ddx[g(x)][g(x)]2 (1)

Calculation:

Obtain the derivative of f(x).

Substitute tanx1 for f(x) and secx for g(x) in equation (1),

ddx[tanx1secx]=[secx]ddx[tanx1][tanx1]ddx[secx][secx]2 =(secx)(sec2x)(tanx1)(secxtanx)(secx)2=sec3x(tanxsecxtanxsecxtanx)(secx)2=sec3x(secxtan2xsecxtanx)(secx)2

Simplify further and obtain the derivative of f(x).

ddx[tanx1secx]=sec3xsecxtan2x+secxtanx(secx)2=secx(sec2xtan2x+tanx)(secx)2=1+tanxsecx     (Qsec2xtan2x=1)=1+tanxsecx

Therefore, the derivative of f(x) is 1+tanxsecx_.

(b)

To determine

To simplify: The expression of f(x) in terms of sinx and cosx. and compute the derivative of f(x).

Expert Solution

Answer to Problem 29E

The function f(x) in terms of sinx and cosx is sinxcosx_.

The derivative of f(x) in terms of sinx and cosx is cosx+sinx_.

Explanation of Solution

Given:

The function is f(x)=tanx1secx.

Difference Rule:

If g1(x) and g2(x) are function, then the differentiation form is,

ddx[g1(x)g2(x)]=ddx[g1(x)]ddx[g2(x)] (2)

Calculation:

Express the function f(x) in terms of sinx and cosx by using trigonometric identities as follows.

f(x)=sinxcosx11cosx      (Q tanx=sinxcosx,secx=1cosx)=sinxcosxcosx1cosx=sinxcosxcosx×cosx1=sinxcosx

Thus, the function f(x) in terms of sinx and cosx is f(x)=sinxcosx_.

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

Apply Difference Rule as shown in equation (2),

ddx[sinxcosx]=ddx[sinx]ddx[cosx]=[cosx][sinx]=cosx+sinx

Thus, the derivative of f(x)=sinxcosx is cosx+sinx_.

(c)

To determine

To show: The answers of parts (a) and (b) are equivalent. That is,1+tanxsecx=cosx+sinx.

Expert Solution

Explanation of Solution

Proof:

From part (a), the function f(x)=1+tanxsecx and from part (b), the function f(x)=cosx+sinx.

Consider f(x)=1+tanxsecx   (From part(a)).

1+tanxsecx=1+sinxcosx1cosx        (Q  tanx=sinxcosx     secx=1cosx)=cosx+sinxcosx1cosx=cosx+sinxcosxcosx=cosx+sinx         (From part(b))

Hence, the solution of parts (a) and (b) are equivalent.

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