Question

Let $y={\tan }^{-1}\left(\frac{3x}{1-2x^2}\right)\forall x\in (0,\frac{1}{\sqrt{2}}),$ then $\frac{dy}{dx}$ at $x=0$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

Given : $y={\tan }^{-1}\left(\frac{3x}{1-2x^2}\right),x\in (0,\frac{1}{\sqrt{2}})$
$\Rightarrow y={\tan }^{-1}\left(\frac{2x+x}{1-2x\cdot x}\right)$
$\Rightarrow y={\tan }^{-1}2x+{\tan }^{-1}x$
$(\because 2x>0,x>0\text{and}2x^2<1)$
$\Rightarrow \frac{dy}{dx}=\frac{2}{1+4x^2}+\frac{1}{1+x^2}$
At $x=0,\frac{dy}{dx}=3$