Question

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

• A. $\frac{dy}{dx}=0$
• B. $\frac{d^{2}y}{d{x}^2}=0$
• C. $\frac{dy}{dx}=\frac{2}{x\left(1+{\ln }^{2}x\right)}$
• D. $\frac{dy}{dx}=1$

Answer 1

The correct option is A $\frac{dy}{dx}=0$

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

${\tan }^{-1}\left(\frac{\ln \frac{e}{x^2}}{\ln e{x}^2}\right)={\tan }^{-1}\left(\frac{1-2\ln x}{1+2\ln x}\right)$

Let $\tan {\theta }_1=1,\tan {\theta }_2=2\ln x$

$={\tan }^{-1}\left(\frac{\tan {\theta }_1-\tan {\theta }_2}{1+\tan {\theta }_1\tan {\theta }_2}\right)$

$={\tan }^{-1}\left(\tan ({\theta }_1-{\theta }_2)\right)$

$=({\theta }_1-{\theta }_2)+n\pi$

${\tan }^{-1}\left(\frac{3+2\ln x}{1-6\ln x}\right)={\tan }^{-1}\left(\frac{3+2\ln x}{1-\left(3\right)\left(2\right)\ln x}\right)$

Let $\tan {\theta }_3=3$

$={\tan }^{-1}\left(\frac{\tan {\theta }_2-\tan {\theta }_2}{1+\tan {\theta }_3\tan {\theta }_2}\right)$

$={\tan }^{-1}\left(\tan ({\theta }_2+{\theta }_3)\right)$

$=({\theta }_2+{\theta }_3)+k\pi {\theta }_3ϵ(0,\frac{\pi }{2});kϵz$

$\therefore y=({\theta }_1-{\theta }_2)+n\pi +({\theta }_2+{\theta }_3)+k\pi$

$={\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(3\right)+(n+k)\pi$

As y is constant, $\therefore \frac{dy}{dx}=0$