Question

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

• A. $\frac{1}{10}log2$
• B. $\frac{1}{5}log2$
• C. $-\frac{1}{10}log2$
• D. $log2$

Answer 1

The correct option is D $log2$

Given,

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

or, $y={\tan }^{-1}\left(\frac{{2.2}^x}{1+(2^x{)}^2}\right)$

or, $y=2{\tan }^{-1}\left(2^x\right)$

Now, $\frac{dy}{dx}=\frac{2}{1+(2^x{)}^2}\times 2^x\times \log 2$.

And ${\frac{dy}{dx}|}_{x=0}=\frac{2}{1+1}\times \log 2=\log 2$.