Question

If $y={\tan }^{-1}\left(\frac{2^{x^{\prime }}}{1+2^{2x+1}}\right)$ 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 C $-\frac{1}{10}log2$

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

$\frac{dy}{dx}$ at $x=0$

$\frac{1}{1+\frac{2^{2x^2}}{(1+2^{2x+1}{)}^2}}.\frac{2\times 2^{x^2}\log 2(1+2^{2x+1})-2^x\left({2.2}^{2x+1}\log 2\right)}{(1+2^{2x+1}{)}^2}$

$\frac{(1+2^{2x+1}{)}^2}{1+2^{4x+2}+{2.2}^{2x+1}+2^{2x^2}}\times \frac{x^2(2^x\log 2+2^x.2{)}^{x+1}-2^x{.2}^{x+1}\log 1}{(1+2^{2+x1{)}^2}}$

$\frac{-\log 2}{1+2^2+2.2+1}=\frac{-1\log 2}{10}$