$\frac{d}{dx}\ \begin{Bmatrix} an^{- 1}{\left( \frac{2x}{1 - x^{2}} ight) + an^{- 1}\left( \frac{3x - x^{3}}{1 - 3x^{2}} ight)} \ - an^{- 1}\left( \frac{4x - 4x^{3}}{1 - 6x^{2} + x^{4}} ight) \end{Bmatrix}$ is equal to

Question

$\frac{d}{dx}\ \begin{Bmatrix} 	an^{- 1}{\left( \frac{2x}{1 - x^{2}} ight) + 	an^{- 1}\left( \frac{3x - x^{3}}{1 - 3x^{2}} ight)} \ - 	an^{- 1}\left( \frac{4x - 4x^{3}}{1 - 6x^{2} + x^{4}} ight) \end{Bmatrix}$ is equal to

Dr. Abubacker · Accepted Answer

(c)
Let $I = \frac{d}{dx}\begin{Bmatrix}
	an^{- 1}{\left( \frac{2x}{1 - x^{2}} ight) + 	an^{- 1}\left( \frac{3x - x^{3}}{1 - 3x^{2}} ight)} \
 - 	an^{- 1}\left( \frac{4x - 4x^{3}}{1 - 6x^{2} + x^{4}} ight)
\end{Bmatrix}$
Put $x = 	an{	heta\ }$the given equation
$	herefore\ \ \ \ I = \frac{d}{dx}\{	an^{- 1}{(	an{2	heta) + 	an^{- 1}{(	an{3	heta) - 	an^{- 1}{(	an{4	heta)\}}}}}}}$
$= \frac{d}{dx}(	heta) = \frac{d}{dx}(	an^{- 1}{x) = \frac{1}{1 + x^{2}}}$

Q. $\frac{d}{dx}\ \begin{Bmatrix} \tan^{- 1}{\left( \frac{2x}{1 - x^{2}} \right) + \tan^{- 1}\left( \frac{3x - x^{3}}{1 - 3x^{2}} \right)} \\ - \tan^{- 1}\left( \frac{4x - 4x^{3}}{1 - 6x^{2} + x^{4}} \right) \end{Bmatrix}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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