Question

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

• A. $\frac{1}{10}\log 2$
• B. $\frac{1}{5}\log 2$
• C. -$\frac{1}{10}\log 2$
• D. $\log 2$

Answer 1

The correct option is C -$\frac{1}{10}\log 2$

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

Differentiate w.r.t. x, we get,

$\frac{dy}{dx}=\frac{d}{dx}\left[{tan}^{-1}\left(\frac{2^x}{1+2^{2x+1}}\right)\right]$

We know that, $\frac{d}{dx}\left[{tan}^{-1}x\right]=\frac{1}{1+x^2}$

$\therefore \frac{dy}{dx}=\frac{1}{1+{\left(\frac{2^x}{1+2^{2x+1}}\right)}^2}\times \frac{d}{dx}\left[\left(\frac{2^x}{1+2^{2x+1}}\right)\right]$

$\therefore \frac{dy}{dx}=\frac{1}{1+{\left(\frac{2^x}{1+2^{2x+1}}\right)}^2}\times \frac{(1+2^{2x+1})\frac{d}{dx}\left(2^x\right)-\left(2^x\right)\frac{d}{dx}(1+2^{2x+1})}{{(1+2^{2x+1})}^2}$

$\therefore \frac{dy}{dx}=\frac{1}{1+{\left(\frac{2^x}{1+2^{2x+1}}\right)}^2}\times \frac{(1+2^{2x+1})2^{x}log2-\left(2^x\right)[2^{2x+1}log2\times \frac{d}{dx}(2x+1)]}{{(1+2^{2x+1})}^2}$

$\therefore \frac{dy}{dx}=\frac{1}{1+{\left(\frac{2^x}{1+2^{2x+1}}\right)}^2}\times \frac{(1+2^{2x+1})2^{x}log2-2^{3x+1}log2\times 2}{{(1+2^{2x+1})}^2}$

At $x=0$,

$\frac{dy}{dx}=\frac{1}{1+{\left(\frac{2^0}{1+2^{0+1}}\right)}^2}\times \frac{(1+2^{0+1})2^{0}log2-2^{0+1}log2\times 2}{{(1+2^{0+1})}^2}$

$\therefore \frac{dy}{dx}=\frac{1}{1+{\left(\frac{1}{1+2^1}\right)}^2}\times \frac{(1+2^1)log2-2^{1}log2\times 2}{{(1+2^1)}^2}$

$\therefore \frac{dy}{dx}=\frac{1}{1+{\left(\frac{1}{3}\right)}^2}\times \frac{3log2-4log2}{{\left(3\right)}^2}$

$\therefore \frac{dy}{dx}=\frac{1}{1+\frac{1}{9}}\times \frac{-log2}{9}$

$\therefore \frac{dy}{dx}=\frac{1}{\frac{10}{9}}\times \frac{-log2}{9}$

$\therefore \frac{dy}{dx}=\frac{9}{10}\times \frac{-log2}{9}$

$\therefore \frac{dy}{dx}=\frac{-log2}{10}$