Question

Differentiate
x+cosx/tanx with respect to x.​

Answer 1

Step-by-step explanation:

We have,

$y = \frac{x + \cos(x) }{ \tan(x) }$

$\implies \: \frac{dy}{dx} = \frac{ \tan(x) \frac{d}{dx}( x + \cos(x) ) - (x + \cos(x)) \frac{d}{dx}( \tan(x) ) }{ \tan^{2} (x) }$

$\implies \: \frac{dy}{dx} = \frac{ \tan(x) ( 1 - \sin(x) ) - (x + \cos(x)) \sec ^{2} (x) }{ \tan^{2} (x) }$

$\implies \: \frac{dy}{dx} = \frac{ \tan(x) - \tan(x) \sin(x) - x \sec ^{2} (x) + \sec(x) }{ \tan^{2} (x) }$

$\implies \: \frac{dy}{dx} = \frac{ \frac{ \sin(x) }{ \cos(x) } - \frac{ \sin^{2} (x)}{ \cos(x) } - \frac{ x}{ \cos ^{2} (x)} + \frac{1}{ \cos(x)} }{ \tan^{2} (x) }$

$\implies \: \frac{dy}{dx} = \frac{ \frac{ \sin(x) \cos(x) - \sin^{2} (x) \cos(x) - x + \cos(x) }{ \cos^{2} (x) } }{ \tan^{2} (x) }$

$\implies \: \frac{dy}{dx} = \frac{ \sin(x) \cos(x) - \sin^{2} (x) \cos(x) - x + \cos(x) }{ \sin^{2} (x) }$

$\implies \frac{dy}{dx} = \cot(x) - \cos(x) - x \cosec ^{2} (x) + \cot(x) \cosec(x)$