The derivative of $\sin^{- 1}\left( \frac{2x}{1 + x^{2}} ight)$ with respect to $ an^{- 1}\left( \frac{2x}{1 - x^{2}} ight)$ is

Question

The derivative of $\sin^{- 1}\left( \frac{2x}{1 + x^{2}} ight)$ with respect to $	an^{- 1}\left( \frac{2x}{1 - x^{2}} ight)$ is

Dr. Abubacker · Accepted Answer

(b)
Let $u = \sin^{- 1}{\left( \frac{2x}{1 + x^{2}} ight) = 2	an^{- 1}x}$
$	herefore\frac{du}{dx} = \frac{2}{1 + x^{2}}$
and $v = 	an^{- 1}{\left( \frac{2x}{1 - x^{2}} ight) = 2	an^{- 1}x}$
$	herefore\frac{dv}{dx} = \frac{2}{1 + x^{2}}$
Hence, $\frac{du}{dv} = \frac{\left( \frac{du}{dx} ight)}{\left( \frac{dv}{dx} ight)} = 1$

Q. The derivative of $\sin^{- 1}\left( \frac{2x}{1 + x^{2}} \right)$ with respect to $\tan^{- 1}\left( \frac{2x}{1 - x^{2}} \right)$ is

Explanation

Reviewed by Dr. Abubacker

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