Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.4 Derivatives of Trigonometric Functions - 2.4 Exercises - Page 151: 31

Answer

$$f(x)=\frac{\tan x-1}{\sec x}$$ (a) By using the Quotient Rule to differentiate the function $f$ we have : $$ \begin{aligned} f^{\prime}(x)&=\frac{\sec x\left(\sec ^{2} x\right)-(\tan x-1)(\sec x \tan x)}{(\sec x)^{2}}\\ &=\frac{\sec x\left(\sec ^{2} x-\tan ^{2} x+\tan x\right)}{\sec ^{2} x}\\ &=\frac{1+\tan x}{\sec x} \end{aligned} $$ (b) Simplify the expression for $f(x)$ by writing it in terms of $\sin x$ and $\cos x,$ and then find $f^{\prime}(x)$ $$ f(x)=\frac{\tan x-1}{\sec x}=\frac{\frac{\sin x}{\cos x}-1}{\frac{1}{\cos x}}=\frac{\frac{\sin x-\cos x}{\cos x}}{\frac{1}{\cos x}}=\sin x-\cos x $$ $\Rightarrow$ $$ f^{\prime}(x)=\cos x-(-\sin x)=\cos x+\sin x$$ $$\text { (c) From part (a), } f^{\prime}(x)=\frac{1+\tan x}{\sec x}=\frac{1}{\sec x}+\frac{\tan x}{\sec x}=\cos x+\sin x, \text { which is the expression for } f^{\prime}(x) \text { in part (b) }$$ So, we find that , the expressions for $ f^{\prime}(x)$in parts (a) and (b) are equivalent.

Work Step by Step

$$f(x)=\frac{\tan x-1}{\sec x}$$ (a) By using the Quotient Rule to differentiate the function $f$ we have : $$ \begin{aligned} f^{\prime}(x)&=\frac{\sec x\left(\sec ^{2} x\right)-(\tan x-1)(\sec x \tan x)}{(\sec x)^{2}}\\ &=\frac{\sec x\left(\sec ^{2} x-\tan ^{2} x+\tan x\right)}{\sec ^{2} x}\\ &=\frac{1+\tan x}{\sec x} \end{aligned} $$ (b) Simplify the expression for $f(x)$ by writing it in terms of $\sin x$ and $\cos x,$ and then find $f^{\prime}(x)$ $$ f(x)=\frac{\tan x-1}{\sec x}=\frac{\frac{\sin x}{\cos x}-1}{\frac{1}{\cos x}}=\frac{\frac{\sin x-\cos x}{\cos x}}{\frac{1}{\cos x}}=\sin x-\cos x $$ $\Rightarrow$ $$ f^{\prime}(x)=\cos x-(-\sin x)=\cos x+\sin x$$ $$\text { (c) From part (a), } f^{\prime}(x)=\frac{1+\tan x}{\sec x}=\frac{1}{\sec x}+\frac{\tan x}{\sec x}=\cos x+\sin x, \text { which is the expression for } f^{\prime}(x) \text { in part (b) }$$ So, the expressions for $ f^{\prime}(x)$in parts (a) and (b) are equivalent.
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