Question

If $y={\tan }^{-1}\left(\frac{2^x}{1+2^{2x+1}}\right)$, then $\frac{dy}{dx}$ at $x=0$ is

• A. 1
• B. 2
• C. $In2$
• D. $\frac{-1}{10}In2$

Answer 1

The correct option is D

$\frac{-1}{10}In2$

Given: $y={\tan }^{-1}\left(\frac{2^x}{1+2^{2x+1}}\right)$

It can be written as :

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

$={\tan }^{-1}{2}^{x+1}-{\tan }^{-1}{2}^x$ $[\because {\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\left(\frac{A-B}{1+AB}\right)]$

Differentiating :

$y^{\prime }=\frac{2^{x+1}ln2}{1+(2^{x+1}{)}^2}-\frac{2^{x}ln2}{1+2^{2x}}$
$\Rightarrow y^{\prime }\left(0\right)=\frac{-1}{10}ln2$