# The derivative of f ( x ) = tan x − 1 sec x .

### Single Variable Calculus: Concepts...

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

### 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)=tanx−1secx.

Expert Solution

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

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

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.

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!