Find the derivative of the following function:
$\left( {x + \cos x} \right)\left( {x - \tan x} \right)$
Answer
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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 cos x, tan x and x which is given as $\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x{\text{ and }}\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x{\text{ and }}\dfrac{d}{{dx}}x = 1$ so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Let
$y = \left( {x + \cos x} \right)\left( {x - \tan x} \right)$
Now differentiate it w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {\left( {x + \cos 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 + \cos x} \right)\dfrac{d}{{dx}}\left( {x - \tan x} \right) + \left( {x - \tan x} \right)\dfrac{d}{{dx}}\left( {x + \cos x} \right)$
Now as we know that differentiation of $\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x{\text{ and }}\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x{\text{ and }}\dfrac{d}{{dx}}x = 1$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {x + \cos x} \right)\left( {1 - {{\sec }^2}x} \right) + \left( {x - \tan x} \right)\left( {1 - \sin x} \right)$
Now simplify it we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {x + \cos x} \right) - {\sec ^2}x\left( {x + \cos x} \right) + \left( {x - \tan x} \right) - \left( {x - \tan x} \right)\sin x$
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {x + \cos x} \right) - x{\sec ^2}x - \sec x + \left( {x - \tan x} \right) - x\sin x + \tan x\sin x$
Now take x coefficients together so we have,
\[ \Rightarrow \dfrac{d}{{dx}}y = x\left( {2 - \sin x - {{\sec }^2}x} \right) + \cos x - \sec x - \tan x + \tan x\sin x\]
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 cos 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 cos x, tan x and x as above and simplify we will get the required answer.
Complete step-by-step answer:
Let
$y = \left( {x + \cos x} \right)\left( {x - \tan x} \right)$
Now differentiate it w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {\left( {x + \cos 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 + \cos x} \right)\dfrac{d}{{dx}}\left( {x - \tan x} \right) + \left( {x - \tan x} \right)\dfrac{d}{{dx}}\left( {x + \cos x} \right)$
Now as we know that differentiation of $\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x{\text{ and }}\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x{\text{ and }}\dfrac{d}{{dx}}x = 1$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {x + \cos x} \right)\left( {1 - {{\sec }^2}x} \right) + \left( {x - \tan x} \right)\left( {1 - \sin x} \right)$
Now simplify it we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {x + \cos x} \right) - {\sec ^2}x\left( {x + \cos x} \right) + \left( {x - \tan x} \right) - \left( {x - \tan x} \right)\sin x$
$ \Rightarrow \dfrac{d}{{dx}}y = \left( {x + \cos x} \right) - x{\sec ^2}x - \sec x + \left( {x - \tan x} \right) - x\sin x + \tan x\sin x$
Now take x coefficients together so we have,
\[ \Rightarrow \dfrac{d}{{dx}}y = x\left( {2 - \sin x - {{\sec }^2}x} \right) + \cos x - \sec x - \tan x + \tan x\sin x\]
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 cos 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 cos x, tan x and x as above and simplify we will get the required answer.
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