#### Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question Question 10

(c)

Hint:

We must know about the derivative of logarithm and  $\tan x$ .

Given:

$y=\tan ^{-1}\left\{\frac{\log _{e}\left(\frac{e}{x^{2}}\right)}{\log _{e}\left(e x^{2}\right)}+\tan ^{-1}\left(\frac{3+2 \log _{e} x}{1-6 \log _{e} x}\right)\right\}$

Explanation:

$y=\tan ^{-1}\left\{\frac{\log _{e}\left(\frac{e}{x^{2}}\right)}{\log _{e}\left(e x^{2}\right)}+\tan ^{-1}\left(\frac{3+2 \log _{e} x}{1-6 \log _{e} x}\right)\right\}$

\begin{aligned} &y=\tan ^{-1}\left(\frac{1-2 \log _{e} x}{1+2 \log _{e} x}\right)+\tan ^{-1}\left(\frac{3+2 \log _{e} x}{1-6 \log _{e} x}\right) \\ & \end{aligned}

$y=\tan ^{-1}\left(\frac{\left(\frac{1-2 \log _{e} x}{1+2 \log _{e} x}\right)+\left(\frac{3+2 \log _{e} x}{1-6 \log _{e} x}\right)}{1-\left(\frac{1-2 \log _{e} x}{1+2 \log _{e} x}\right)\left(\frac{3+2 \log _{e} x}{1-6 \log _{e} x}\right)}\right)$

\begin{aligned} &y=\tan ^{-1}\left\{\frac{\left(1-2 \log _{e} x\right)\left(1-6 \log _{e} x\right)+\left(3+2 \log _{e} x\right)\left(1+2 \log _{e} x\right)}{\left(1+2 \log _{e} x\right)\left(1-6 \log _{e} x\right)-\left(1-2 \log _{e} x\right)\left(3+2 \log _{e} x\right)}\right\} \\ & \end{aligned}

$y=\tan ^{-1}\left\{\frac{1-8 \log _{e} x+12\left(\log _{e} x\right)^{2}+3+8 \log _{e} x+4\left(\log _{e} x\right)^{2}}{1-4 \log _{e} x-12\left(\log _{e} x\right)^{2}-3+4 \log _{e} x+4\left(\log _{e} x\right)^{2}}\right\}$

\begin{aligned} &y=\tan ^{-1}\left\{\frac{4+16\left(\log _{e} x\right)^{2}}{-2-8\left(\log _{e} x\right)^{2}}\right\} \\ & \end{aligned}

$y=\tan ^{-1}\left\{\frac{4\left(1+4\left(\log _{e} x\right)^{2}\right)}{-2\left(1+4\left(\log _{e} x\right)^{2}\right)}\right\}$

\begin{aligned} &y=\tan ^{-1}[-2] \\ & \end{aligned}

$\frac{d y}{d x}=0, \frac{d^{2} y}{d x^{2}}=0$