Question

Differentiate with respect to x :
tan^-1 ( 2x / 1 + 15x² )
The answer: 5 /( 1 + 25x² ) - 3 / ( 1 + 9x² )

Answer 1

$$\begin{gathered} \mathrm{Let}y={\tan }^{-1}\left[\frac{2x}{1+15x^2}\right] \\ \Rightarrow y={\tan }^{-1}\left[\frac{5x-3x}{1+5x.3x}\right] \\ \mathrm{put}5x=\tan \alpha ;3x=\tan \beta \\ \mathrm{Now},y={\tan }^{-1}\left[\frac{\tan \alpha -\tan \beta }{1+\tan \alpha .\tan \beta }\right] \\ \Rightarrow y={\tan }^{-1}\left[\tan \left(\alpha -\beta \right)\right] \\ \Rightarrow y=\alpha -\beta \\ \Rightarrow y={\tan }^{-1}5x-{\tan }^{-1}3x \\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+25x^2}\times 5-\frac{1}{1+9x^2}\times 3 \\ \Rightarrow \boxed{\frac{dy}{dx}=\frac{5}{1+25x^2}-\frac{3}{1+9x^2}} \end{gathered}$$