If $y = an^{- 1}{\left( \frac{\log{(e/x^{2})}}{\log{(ex^{2})}} ight) + an^{- 1}\left( \frac{3 + 2\log x}{1 - 6\log x} ight)}$, then $\frac{d^{2}y}{dx^{2}}$ is equal to

Question

If $y = 	an^{- 1}{\left( \frac{\log{(e/x^{2})}}{\log{(ex^{2})}} ight) + 	an^{- 1}\left( \frac{3 + 2\log x}{1 - 6\log x} ight)}$, then $\frac{d^{2}y}{dx^{2}}$ is equal to

Dr. Abubacker · Accepted Answer

(c)
$\because y = 	an^{- 1}{\left( \frac{\log\left( \frac{e}{x^{2}} ight)}{\log{ex^{2}}} ight) + 	an^{- 1}\left( \frac{3 + 2\log x}{1 - 6\log x} ight)}$
$= 	an^{- 1}\left( \frac{1 - \log x^{2}}{1 + \log x^{2}} ight) + 	an^{- 1}\left( \frac{3 + 2\log x}{1 - 6\log x} ight)$
$= 	an^{- 1}{(1) - 	an^{- 1}{(2\log{x) + 	an^{- 1}{(3) + 	an^{- 1}{(2\log{x)}}}}}}$
$	herefore y = 	an^{- 1}{(1) + 	an^{- 1}{(3)}}$
$\Rightarrow \frac{dy}{dx} = 0\  \Rightarrow \ \frac{d^{2}y}{dx^{2}} = 0$

Q. If $y = \tan^{- 1}{\left( \frac{\log{(e/x^{2})}}{\log{(ex^{2})}} \right) + \tan^{- 1}\left( \frac{3 + 2\log x}{1 - 6\log x} \right)}$, then $\frac{d^{2}y}{dx^{2}}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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