⇒Derivative

Correct Option is (D)

$\begin{array}{rl}& f(x)={\sin }^{-1}\left(\frac{2\log x}{1+(\log x{)}^2}\right)\\ \therefore f^{\prime }(x)=& \frac{1}{\sqrt{1-{\left(\frac{2\log x}{1+(\log x{)}^2}\right)}^2}}\end{array}$

$\begin{array}{rl}& =\frac{1+(\log x{)}^2}{\sqrt{1+(\log x{)}^4+2(\log x{)}^2-4(\log x{)}^2}}\times \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{2\log x}{1+(\log x{)}^2}\right)\\ & =\frac{1+(\log x{)}^2}{\sqrt{1-2(\log x{)}^2+(\log x{)}^4}}\\ & =\frac{1}{1-(\log x{)}^2}\times \frac{2+2(\log x{)}^2-4(\log x{)}^2}{x[1+(\log x{)}^2]}\\ =& \frac{1}{1-(\log x{)}^2]\times \frac{2}{x}-(2\log x)\left(\frac{2\log x}{x}\right)}\times \frac{2[1-(\log x{)}^2]}{x[1+(\log x{)}^2]}\\ =& \frac{2}{x[1+(\log x{)}^2]}\\ \therefore {\mathrm{f}}^{\prime }(\mathrm{e})& =\frac{1}{\mathrm{e}}\end{array}$

Correct Option is (A)

$\begin{array}{rl}& y={\tan }^{-1}\left(\frac{\log \left(\frac{\mathrm{e}}{x^2}\right)}{\log \left(e{x}^2\right)}\right)+{\tan }^{-1}\left(\frac{4+2\log x}{1-8\log x}\right)\\ & ={\tan }^{-1}\left(\frac{\log e-\log x^2}{\log e+\log x^2}\right)+{\tan }^{-1}(4)+{\tan }^{-1}(2\log x)\\ & ={\tan }^{-1}\left(\frac{1-2\log x}{1+2\log x}\right)+{\tan }^{-1}(4)+{\tan }^{-1}(2\log x)\\ & ={\tan }^{-1}(1)-{\tan }^{-1}(2\log x)+{\tan }^{-1}(4)+{\tan }^{-1}(2\log x)\\ & \therefore y={\tan }^{-1}(1)+{\tan }^{-1}(4)\\ & \therefore \frac{dy}{dx}=0\end{array}$

Correct Option is (B)

$(2x{)}^{2y}=4{\mathrm{e}}^{2x-2y}$

Taking log on both sides, we get

$\begin{array}{rl}& 2y\log 2x=\log \left(4{\mathrm{e}}^{2x-2y}\right)\\ & \Rightarrow 2y\log 2x=\log 4+\log {\mathrm{e}}^{2x-2y}\\ & \Rightarrow 2y\log 2x=\log 4+2x-2y\\ & \Rightarrow 2y\log 2x+2y=\log 4+2x\\ & \Rightarrow 2(y\log 2x+y)=2\log 2+2x\\ & \Rightarrow y\log 2x+y=\log 2+x\\ & \Rightarrow y(1+\log 2x)=x+\log 2\\ & \Rightarrow y=\frac{x+\log 2}{1+\log 2x}\end{array}$

Differentiating w.r.t. $x$, we get

$\begin{array}{rl}& \frac{dy}{dx}=\frac{(1+\log 2x)(1+0)-(x+\log 2)\left(\frac{1}{2x}\right)\cdot 2}{(1+\log 2x{)}^2}\\ & \Rightarrow \frac{dy}{dx}=\frac{(1+\log 2x)-\frac{1}{x}(x+\log 2)}{(1+\log 2x{)}^2}\\ & \Rightarrow (1+\log 2x{)}^2\frac{dy}{dx}=1+\log 2x-1-\frac{\log 2}{x}\\ & \Rightarrow (1+\log 2x{)}^2\frac{dy}{dx}=\log 2x-\frac{\log 2}{x}\\ & =\frac{x\log 2x-\log 2}{x}\end{array}$

Correct Option is (B)

$\begin{array}{ll}& \mathrm{g}(x)\text{ is inverse of function }\mathrm{f}(x)\\ & \text{ i.e., }\mathrm{g}(x)={\mathrm{f}}^{-1}(x)\\ & \mathrm{f}(\mathrm{g}(x))=x\\ & \text{ differentiating w.r.t. }x\text{, we get }\\ & {\mathrm{f}}^{\prime }(\mathrm{g}(x))\times {\mathrm{g}}^{\prime }(x)=1\\ \therefore & {\mathrm{g}}^{\prime }(x)=\frac{1}{{\mathrm{f}}^{\prime }(\mathrm{g}(x))}\text{.... (i)}\\ & \text{ Now, }{\mathrm{f}}^{\prime }(x)=\frac{1}{1+x^3}\text{....[Given]}\\ \therefore & {\mathrm{f}}^{\prime }(\mathrm{g}(x))=\frac{1}{1+(\mathrm{g}(x){)}^3}\dots \text{ (ii) }\\ \therefore & \text{ From (ii) and }(\text{ii}),\text{ we get }\\ \therefore & {\mathrm{g}}^{\prime }(x)=1+(\mathrm{g}(x){)}^3\end{array}$

Correct Option is (C)

$y={\tan }^{-1}\left\{\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\right\}$

Put $a=r\cos \alpha ,b=r\sin \alpha$

$\begin{array}{rl}\therefore y& ={\tan }^{-1}\left\{\frac{\mathrm{r}(\cos \mathrm{x}\cos \alpha -\sin \mathrm{x}\sin \alpha )}{\mathrm{r}(\sin \alpha \cos \mathrm{x}+\cos \alpha \sin \mathrm{x})}\right\}\\ & ={\tan }^{-1}\left[\frac{\cos (\mathrm{x}+\alpha )}{\sin (\mathrm{x}+\alpha )}\right]={\tan }^{-1}[\cot (\mathrm{x}+\alpha )]\\ & ={\tan }^{-1}\{\tan [\frac{\pi }{2}-(\mathrm{x}+\alpha )]\}=\frac{\pi }{2}-(\mathrm{x}+\alpha )\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}& =0-(1+0)=-1\end{array}$