Question

If $y={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)-{\tan }^1\left(\frac{4x-4x^3}{1-6x^2+x^4}\right)$ then Show that $\frac{dy}{dx}=\frac{1}{1+x^2}$

Answer 1

$\begin{array}{c}y={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\tan }^{-1}\left(\frac{3x-x^3}{13x^2}\right)-{\tan }^{-1}\left(\frac{4x-4x^3}{1-6x^2+x^4}\right)\\ \Rightarrow letx=\tan \theta \\ \Rightarrow and\theta ={\tan }^{-1}x\\ \Rightarrow y={\tan }^{-1}\tan 2\theta +{\tan }^{-1}\tan 3\theta -{\tan }^{-1}\tan 4\theta \\ \Rightarrow y=2\theta +3\theta -4\theta \\ \Rightarrow y=\theta ={\tan }^{-1}x\end{array}$

On differentiating w.r t $x$, we get

$\begin{array}{c}\Rightarrow \frac{dy}{dx}=\frac{d\left({\tan }^{-1}x\right)}{dx}\\ \frac{dy}{dx}=\frac{1}{\left(1+x^2\right)}\end{array}$