Chapter 3.3, Problem 29E

(a)

To determine

To find: The derivative of $f\left(x\right)=\frac{\tan x-1}{\sec x}$.

Answer to Problem 29E

The derivative of $f\left(x\right)$ is $\underset{\_}{\frac{1+\tan x}{\sec x}}$.

Explanation of Solution

Given:

The function is $f\left(x\right)=\frac{\tan x-1}{\sec x}$.

Derivative rule: Quotient rule

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

$\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]-f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]}{{\left[g\left(x\right)\right]}^2}$ (1)

Calculation:

Obtain the derivative of $f\left(x\right)$.

Substitute $\tan x-1$ for $f\left(x\right)$ and $\sec x$ for $g\left(x\right)$ in equation (1),

$\begin{array}{c}\frac{d}{dx}\left[\frac{\tan x-1}{\sec x}\right]=\frac{\left[\sec x\right]\frac{d}{dx}\left[\tan x-1\right]-\left[\tan x-1\right]\frac{d}{dx}\left[\sec x\right]}{{\left[\sec x\right]}^2}\\ \text{=}\frac{\left(\sec x\right)\left({\sec }^{2}x\right)-\left(\tan x-1\right)\left(\sec x\tan x\right)}{{\left(\sec x\right)}^2}\\ =\frac{{\sec }^{3}x-\left(\tan x\sec x\tan x-\sec x\tan x\right)}{{\left(\sec x\right)}^2}\\ =\frac{{\sec }^{3}x-\left(\sec x{\tan }^{2}x-\sec x\tan x\right)}{{\left(\sec x\right)}^2}\end{array}$

Simplify further and obtain the derivative of $f\left(x\right)$.

$\begin{array}{c}\frac{d}{dx}\left[\frac{\tan x-1}{\sec x}\right]=\frac{{\sec }^{3}x-\sec x{\tan }^{2}x+\sec x\tan x}{{\left(\sec x\right)}^2}\\ =\frac{\sec x\left({\sec }^{2}x-{\tan }^{2}x+\tan x\right)}{{\left(\sec x\right)}^2}\\ =\frac{1+\tan x}{\sec x}\left(Q{\sec }^{2}x-{\tan }^{2}x=1\right)\\ =\frac{1+\tan x}{\sec x}\end{array}$

Therefore, the derivative of $f\left(x\right)$ is $\underset{\_}{\frac{1+\tan x}{\sec x}}$.

(b)

To determine

To simplify: The expression of $f\left(x\right)$ in terms of $\sin x$ and $\cos x$. and compute the derivative of $f\left(x\right)$.

Answer to Problem 29E

The function $f\left(x\right)$ in terms of $\sin x$ and $\cos x$ is $\underset{\_}{\sin x-\cos x}$.

The derivative of $f\left(x\right)$ in terms of $\sin x$ and $\cos x$ is $\underset{\_}{\cos x+\sin x}$.

Explanation of Solution

Given:

The function is $f\left(x\right)=\frac{\tan x-1}{\sec x}$.

Difference Rule:

If $g_1\left(x\right)$ and $g_2\left(x\right)$ are function, then the differentiation form is,

$\frac{d}{dx}\left[g_1\left(x\right)-g_2\left(x\right)\right]=\frac{d}{dx}\left[g_1\left(x\right)\right]-\frac{d}{dx}\left[g_2\left(x\right)\right]$ (2)

Calculation:

Express the function $f\left(x\right)$ in terms of $\sin x$ and $\cos x$ by using trigonometric identities as follows.

$\begin{array}{c}f\left(x\right)=\frac{\frac{\sin x}{\cos x}-1}{\frac{1}{\cos x}}\left(Q\tan x=\frac{\sin x}{\cos x},\sec x=\frac{1}{\cos x}\right)\\ =\frac{\frac{\sin x-\cos x}{\cos x}}{\frac{1}{\cos x}}\\ =\frac{\sin x-\cos x}{\cos x}\times \frac{\cos x}{1}\\ =\sin x-\cos x\end{array}$

Thus, the function $f\left(x\right)$ in terms of $\sin x$ and $\cos x$ is $\underset{\_}{f\left(x\right)=\sin x-\cos x}$.

The derivative of $f\left(x\right)=\sin x-\cos x$ is computed as follows,

Apply Difference Rule as shown in equation (2),

$\begin{array}{c}\frac{d}{dx}\left[\sin x-\cos x\right]=\frac{d}{dx}\left[\sin x\right]-\frac{d}{dx}\left[\cos x\right]\\ =\left[\cos x\right]-\left[-\sin x\right]\\ =\cos x+\sin x\end{array}$

Thus, the derivative of $f\left(x\right)=\sin x-\cos x$ is $\underset{\_}{\cos x+\sin x}$.

(c)

To determine

To show: The answers of parts (a) and (b) are equivalent. That is,$\frac{1+\tan x}{\sec x}=\cos x+\sin x$.

Explanation of Solution

Proof:

From part (a), the function $f^{\prime }\left(x\right)=\frac{1+\tan x}{\sec x}$ and from part (b), the function $f^{\prime }\left(x\right)=\cos x+\sin x$.

Consider $f^{\prime }\left(x\right)=\frac{1+\tan x}{\sec x}\left(\text{From part}\left(a\right)\right)$.

$\begin{array}{c}\frac{1+\tan x}{\sec x}=\frac{1+\frac{\sin x}{\cos x}}{\frac{1}{\cos x}}\left(\begin{array}{l}Q\tan x=\frac{\sin x}{\cos x}\\ \sec x=\frac{1}{\cos x}\end{array}\right)\\ =\frac{\frac{\cos x+\sin x}{\cos x}}{\frac{1}{\cos x}}\\ =\frac{\cos x+\sin x}{\cos x}\cos x\\ =\cos x+\sin x\left(\text{From part}\left(b\right)\right)\end{array}$

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