Question

Differentiate \[x^\left( \sin x - \cos x \right) + \frac{x^2 - 1}{x^2 + 1}\] ?

Answer 1

\[\text{Let y } = x^\left( \sin x - \cos x \right) + \left( \frac{x^2 - 1}{x^2 + 1} \right)\]

\[ \Rightarrow y = e^{\log x^\left( \sin x - \cos x \right)} + \left( \frac{x^2 - 1}{x^2 + 1} \right)\]

\[ \Rightarrow y = e^{ \left( \sin x - \cos x \right)\log x } + \left( \frac{x^2 - 1}{x^2 + 1} \right)\]

Differentiate it with respect to x using chain rule,

\[\frac{dy}{dx} = \frac{d}{dx}\left[ e^{\left( \sin x - \cos x \right)\log x }\right] + \frac{d}{dx}\left[ \frac{x^2 - 1}{x^2 + 1} \right]\] 

\[             =  e^{\left( \sin x - \cos x \right)\log x }\frac{d}{dx}\left\{ \left( \sin x - \cos x \right)\log x \right\} + \left[ \frac{\left( x^2 + 1 \right)\frac{d}{dx}\left( x^2 - 1 \right) - \left( x^2 - 1 \right)\frac{d}{dx}\left( x^2 + 1 \right)}{\left( x^2 + 1 \right)^2} \right]\] 

\[             =  e^{ \log x^\left( \sin x - \cos x  \right) } \left[ \left( \sin x - \cos x \right)\frac{d}{dx}\left( \log x \right) + \left( \log x \right)\frac{d}{dx}\left( \sin x - \cos x \right) \right] + \left[ \frac{\left( x^2 + 1 \right)\left( 2x \right) - \left( x^2 - 1 \right)\left( 2x \right)}{\left( x^2 + 1 \right)^2} \right]\] 

\[             =  x^\left( \sin x - \cos x \right) \left[ \left( \sin x - \cos x \right)\left( \frac{1}{x} \right) + \log x\left( \sin x + \cos x \right) \right] + \left[ \frac{2 x^3 + 2x - 2 x^3 + 2x}{\left( x^2 + 1 \right)^2} \right]\] 

\[             =  x^\left( \sin x - \cos x \right) \left[ \frac{\left( \sin x - \cos x \right)}{x} + \left( \sin x + \cos x \right)\log x \right] + \frac{4x}{\left( x^2 + 1 \right)^2}\]