Question

Find the derivative of the following function:
$\left( {x + \sec x} \right)\left( {x - \tan x} \right)$

Answer 1

Hint: In this question apply the product rule of differentiation which is given as $\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ later on in the solution apply the differentiation property of sec x, tan x and x which is given as $\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x{\text{ and }}\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x{\text{ and }}\dfrac{d}{{dx}}x = 1$ so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Let
$y = \left( {x + \sec x} \right)\left( {x - \tan x} \right)$
Now differentiate it w.r.t. x we have,
$\Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {\left( {x + \sec x} \right)\left( {x - \tan x} \right)} \right]$
Now here we use product rule of differentiate which is given as
$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ so use this property in above equation we have,
$\Rightarrow \dfrac{d}{{dx}}y = \left( {x + \sec x} \right)\dfrac{d}{{dx}}\left( {x - \tan x} \right) + \left( {x - \tan x} \right)\dfrac{d}{{dx}}\left( {x + \sec x} \right)$
Now as we know that differentiation of $\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x{\text{ and }}\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x{\text{ and }}\dfrac{d}{{dx}}x = 1$ so use this property in above equation we have,
$\Rightarrow \dfrac{d}{{dx}}y = \left( {x + \sec x} \right)\left( {1 - {{\sec }^2}x} \right) + \left( {x - \tan x} \right)\left( {1 + \sec x\tan x} \right)$
Now simplify it we have,
$\Rightarrow \dfrac{d}{{dx}}y = \left( {x + \sec x} \right) - {\sec ^2}x\left( {x + \sec x} \right) + \left( {x - \tan x} \right) + \left( {x - \tan x} \right)\sec x\tan x$
$\Rightarrow \dfrac{d}{{dx}}y = \left( {x + \sec x} \right) - x{\sec ^2}x - {\sec ^3}x + \left( {x - \tan x} \right) - x\sec x\tan x - \sec x{\tan ^2}x$
Now take x coefficients together so we have,
$\Rightarrow \dfrac{d}{{dx}}y = x\left( {2 - \sec x\tan x - {{\sec }^2}x} \right) + \sec x - {\sec ^3}x - \tan x - \sec x{\tan ^2}x$
So this is the required differentiation.

Note – Whenever we face such types of questions the key concept is always recall the formula of product rule of differentiation, formula of sec x, tan x and x differentiation which is stated above then first apply the product rule as above then use the property of differentiation of sec x, tan x and x as above and simplify we will get the required answer.