Question

Let $y={\sin }^{-1}\frac{2x}{1+x^2}+{\tan }^{-1}x$. Then find $\frac{dy}{dx}$.

Answer 1

Given,

$y={\sin }^{-1}\frac{2x}{1+x^2}+{\tan }^{-1}x$

or, $y=2{\tan }^{-1}x+{\tan }^{-1}x$ [ Using direct formula of $2{\tan }^{-1}x$]

or, $y=3{\tan }^{-1}x$

Now differentiating both sides with respect to $x$ we get,

$\frac{dy}{dx}=\frac{3}{1+x^2}$.