Question

Find the derivative of ${\sin }^{-1}\frac{2x}{1+x^2}$ with respect to ${\tan }^{-1}\frac{2x}{1-x^2}$.

Answer 1

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

$z={\tan }^{-1}\frac{2x}{1-x^2}=2{\tan }^{-1}x$. [ Since $2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}$ and $2{\tan }^{-1}x={\sin }^{-1}\frac{2x}{1+x^2}$].

So $y=z$......(1).

Now we are to find $\frac{dy}{dz}$.

Now form (1) we have

$\frac{dy}{dz}=1$.