Question

What is the derivative of the function,
$f(x) = {e^{\tan x}} + \ln (\sec x) - {e^{\ln x}}$ at $x = \dfrac{\pi }{4}$
(A) e/2
(B) e
(C) 2e
(D) 4e

Answer 1

Hint: Differentiation of ${e^u}$ with respect to x is ${e^u}.\dfrac{{du}}{{dx}}$ and that of ln(u) with respect to x is $\dfrac{1}{u}.\dfrac{{du}}{{dx}}$. Use these rules to differentiate the above function with respect to x and then put $x = \dfrac{\pi }{4}$ to find the required value.

Complete step by step answer:
From the given question,
$\Rightarrow f(x) = {e^{\tan x}} + \ln (\sec x) - {e^{\ln x}}$
We know that ${e^{\ln x}} = x$. So using this, we’ll get:
$\Rightarrow f(x) = {e^{\tan x}} + \ln (\sec x) - x$
Differentiating it with respect to x, we’ll get:
$\Rightarrow f'(x) = \dfrac{d}{{dx}}{e^{\tan x}} + \dfrac{d}{{dx}}\ln (\sec x) - \dfrac{d}{{dx}}x$
We know that $\dfrac{d}{{dx}}{e^u} = {e^u}.\dfrac{{du}}{{dx}}$, $\dfrac{d}{{dx}}\ln u = \dfrac{1}{u}.\dfrac{{du}}{{dx}}$ and $\dfrac{d}{{dx}}x = 1$. Using these results, we’ll get:
$\Rightarrow f'(x) = {e^{\tan x}}.\dfrac{d}{{dx}}\tan x + \dfrac{1}{{\sec x}}.\dfrac{d}{{dx}}\sec x - 1$
Further, we know that $\dfrac{d}{{dx}}\tan x = {\sec ^2}x$ and $\dfrac{d}{{dx}}\sec x = \sec x\tan x$. Using these formulas, we’ll get:
$\Rightarrow f'(x) = {e^{\tan x}}.{\sec ^2}x + \dfrac{1}{{\sec x}}\sec x\tan x - 1 \\\ \Rightarrow f'(x) = {e^{\tan x}}.{\sec ^2}x + \tan x - 1 \\\$
Now putting $x = \dfrac{\pi }{4}$, we’ll get:
$\Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = {e^{\tan \dfrac{\pi }{4}}}.{\sec ^2}\dfrac{\pi }{4} + \tan \dfrac{\pi }{4} - 1$
We also know that $\tan \dfrac{\pi }{4} = 1$ and $\sec \dfrac{\pi }{4} = \sqrt 2$. Putting these values, we’ll get:
$\Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = {e^1}.{\left( {\sqrt 2 } \right)^2} + 1 - 1 \\\ \Rightarrow f'\left( {\dfrac{\pi }{4}} \right) = 2e \\\$

Therefore (C) is the correct option.

Note: We have used the chain rule of differentiation in the problem. According to this rule, if a function is such that $y = f\left( {g\left( x \right)} \right)$, then its differentiation is:
$\Rightarrow \dfrac{{dy}}{{dx}} = f'\left( {g\left( x \right)} \right).g'\left( x \right)$
If we have to differentiate a composite function we have to use this method.