Question

Differentiate with respect to x: ${e^{ax}}\sec x\tan 2x$

Answer 1

Hint : Here in this question, we consider the given function as y and we are going to differentiate the given function with respect to x. The function is a product of 3 terms containing x. so to differentiate the function we use product rules and then we are going to simplify.

Complete step-by-step answer :
Now consider the given function as y, so we have
$y = {e^{ax}}\sec x\tan 2x$ -----(1)
Differentiate the function with respect to x. We apply product rule, the product rule is $\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
Therefore, we have
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}({e^{ax}}\sec x\tan 2x)$
Since the given function is in the form of product of 3 terms and the 3 terms are the function of x so it is necessary to apply the product rule for the function y
We apply product rule, so we consider ${e^{ax}}$ as first function and $\sec x\tan 2x$ as the second function
So, we have
$\dfrac{{dy}}{{dx}} = {e^{ax}}\dfrac{d}{{dx}}(\sec x\tan 2x) + \sec x\tan 2x\dfrac{d}{{dx}}({e^{ax}})$
Again, we have to apply product rule to $\sec x\tan 2x$
Here we take $\sec x$ as a first function and $\tan 2x$ as a second function
Therefore, we have
$\dfrac{{dy}}{{dx}} = {e^{ax}}\left[ {\sec x\dfrac{d}{{dx}}(\tan 2x) + \tan 2x\dfrac{d}{{dx}}(\sec x)} \right] + \sec x\tan 2x\dfrac{d}{{dx}}({e^{ax}})$
Applying differentiation to the terms, we have
$\Rightarrow \dfrac{{dy}}{{dx}} = {e^{ax}}\left[ {\sec x.{{\sec }^2}2x.(2) + \tan 2x.\sec x.\tan x} \right] + \sec x\tan 2x.a.{e^{ax}}$
Rearrange the terms, we have
$\Rightarrow \dfrac{{dy}}{{dx}} = 2{e^{ax}}\sec x.{\sec ^2}2x + {e^{ax}}\tan 2x.\sec x.\tan x + a{e^{ax}}\sec x\tan 2x$
Since the $\dfrac{{dy}}{{dx}}$ contains ${e^{ax}}$ in three terms.
We take ${e^{ax}}$ as common and $\dfrac{{dy}}{{dx}}$ is written as
$\Rightarrow \dfrac{{dy}}{{dx}} = {e^{ax}}(2\sec x.{\sec ^2}2x + \tan 2x.\sec x.\tan x + a\sec x\tan 2x)$
Since the $\dfrac{{dy}}{{dx}}$ contains $\sec x$ in three terms.
We take $\sec x$ as common and $\dfrac{{dy}}{{dx}}$ is written as
$\Rightarrow \dfrac{{dy}}{{dx}} = {e^{ax}}\sec x(2{\sec ^2}2x + \tan 2x.\tan x + a\tan 2x)$
Therefore, we have

$\dfrac{d}{{dx}}({e^{ax}}\sec x\tan 2x) = {e^{ax}}\sec x(2{\sec ^2}2x + \tan 2x.\tan x + a\tan 2x)$
Hence, we obtained the required result.
So, the correct answer is “${e^{ax}}\sec x(2{\sec ^2}2x + \tan 2x.\tan x + a\tan 2x)$
”.

Note : The differentiation is defined as the derivative of a function with respect to the independent variable. Here the dependent variable is y and the independent variable is x. If the function is a product of more than one function, we use product rule to find the derivative. The product rule is defined as $\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$ , where u and v are both the function of x. By using the differentiation formulas we can obtain the result.