Question

$Ify={\tan }^{-1}\left[\frac{2^x}{1+2^{x+1}}\right],-\infty <x<\infty then\frac{dy}{dx}atx=0is$

• A. $\frac{-3}{5}\ln 2$
• B. $\frac{1}{10}\ln 2$
• C. $2$
• D. None of these

Answer 1

The correct option is B $\frac{1}{10}\ln 2$

We know that $\frac{d}{dx}\left(ta{n}^{-1}x\right)=\frac{1}{1+x^2}$ , but in this case the argument of $ta{n}^{-1}x$ is a different function of x and not x itself. So we will apply chain rule.
$y=ta{n}^{-1}\frac{2^x}{1+2^{x+1}}$
$\frac{dy}{dx}=\frac{1}{1+{\left[\frac{2^x}{1+2^{x+1}}\right]}^2}\times \frac{d}{dx}\left(\frac{2^x}{1+2^{x+1}}\right)$ [Using chain rule]
$=\frac{1}{1+{\left[\frac{2^x}{1+2^{x+1}}\right]}^2}\times \frac{2^x\ln 2(1+2^{x+1})-2^x\times 2^{x+1}\ln 2}{(1+2^{x+1}{)}^2}$ [Using quotient rule]
${Putx=0toget\frac{dy}{dx}|}_{x=0}=\frac{1}{10}\ln 2$ which is the correct answer