Question

If $y={\tan }^{-1}\left[\frac{\log \left(\frac{e}{x^3}\right)}{\log \left({ex}^3\right)}\right]$ , then $\frac{dy}{dx}=?$

• A. $\frac{-3}{1+9{\left(\log x\right)}^2}$
• B. $\frac{3}{1+9{\left(\log x\right)}^2}$
• C. $\frac{-3}{x(1+9{\left(\log x\right)}^2)}$
• D. $\frac{3}{x(1+9{\left(\log x\right)}^2)}$

Answer 1

The correct option is A $\frac{-3}{1+9{\left(\log x\right)}^2}$

The correct option is C

$y={\tan }^{-1}\left[\frac{\log \left(\frac{e}{x^3}\right)}{\log \left(e{x}^3\right)}\right]$

$\to$ $\tan y=\left[\frac{\log \left(\frac{e}{x^3}\right)}{\log \left(e{x}^3\right)}\right]$

$\to$$\tan y=\left[\frac{\log e-\log x^3}{\log e+\log x^3}\right]$

$\to$$\tan y=\left[\frac{1-3\log x}{1+3\log x}\right]$ - - - - - - (1)

Differentiating both sides with respect to x we get :

${\sec }^{2}y\frac{dy}{dx}=\frac{(1+3\log x)\frac{d}{dx}(1-3\log x)-(1-3\log x)\frac{d}{dx}(1+3\log x)}{(1+3\log x{)}^2}$

$\to$ $(1+{\tan }^{2}y)\frac{dy}{dx}=\frac{(1+3\log x)\left(\frac{-3}{x}\right)-(1-3\log x)\left(\frac{3}{x}\right)}{(1+3\log x{)}^2}$

Using (1) we get :

$[1+{\left(\frac{1-3\log x}{1+3\log x}\right)}^2]\frac{dy}{dx}=\frac{\frac{-6}{x}}{(1+3\log x{)}^2}$

$\to$$\left[\frac{(1+3\log x{)}^2+(1-3\log x{)}^2}{(1+3\log x{)}^2}\right]\frac{dy}{dx}=\frac{\frac{-6}{x}}{(1+3\log x{)}^2}$

$\to$$\left[\frac{2(1+9(\log x{)}^2)}{(1+3\log x{)}^2}\right]\frac{dy}{dx}=\frac{\frac{-6}{x}}{(1+3\log x{)}^2}$

$\to$$\frac{dy}{dx}=\frac{\frac{-6}{x}}{2(1+9(\log x{)}^2)}$

$\to$$\frac{dy}{dx}=\frac{-3}{x(1+9(\log x{)}^2)}$