Question

Differentiate x+ cos x/tan x with respect to x​

Answer 1

Answer:

Here is the answer for your question.

Answer 2

GIVEN :–

• A function $\:\: \bf \dfrac{x + \cos(x)}{\tan(x)}$.

TO FIND :–

• Differentiate with respect to 'x' = ?

SOLUTION :–

• Let –

$\\ \bf \implies P = \dfrac{x + \cos(x)}{\tan(x)}$

• We know that –

$\\ \bf \implies \dfrac{d}{dx} \left(\dfrac{u}{v} \right) = \dfrac{v \dfrac{du}{dx} - u \dfrac{dv}{dx}}{ {v}^{2} }$

• So that –

$\\ \bf \implies \dfrac{dP}{dx} = \dfrac{ \tan(x) \dfrac{d}{dx} \{x + \cos(x) \} + \left\{x + \cos(x)\right\}\dfrac{d \{ \tan(x) \}}{dx} }{\tan^{2} (x)}$

• We also know that –

$\\ \bf \longrightarrow \dfrac{d (\tan x)}{dx} = \sec^{2} (x)$

$\\ \bf \longrightarrow \dfrac{d (\cos x)}{dx} = - \sin(x)$

• So that –

$\\ \bf \implies \dfrac{dP}{dx} = \dfrac{ \tan(x)\{1 -\sin(x) \} + \left\{x + \cos(x)\right\}\sec^{2}(x)}{\tan^{2} (x)}$

$\\ \bf \implies \dfrac{dP}{dx} = \dfrac{ \tan(x)-\sin(x) \tan(x) + x\sec^{2}(x)+ \cos(x)\sec^{2}(x)}{\tan^{2} (x)}$

$\\ \bf \implies \dfrac{dP}{dx} = \dfrac{ \tan(x)-\sin(x) \tan(x) + x\sec^{2}(x)+ \sec(x)}{\tan^{2} (x)}$