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

Question

Dr. Abubacker · Accepted Answer

(b)
Let $p = \sin^{- 1}{\frac{2x}{1 + x^{2}} = 2	an^{- 1}x}$
and $q = \cos^{- 1}{\frac{1 - x^{2}}{1 + x^{2}} = 2	an^{- 1}x}$
$	herefore$ $\ \ p = q\ \ \  \Rightarrow \ \ \ \ \frac{dp}{dq} = 1$

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

Explanation

Reviewed by Dr. Abubacker

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