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.