Question

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

• A. $-\frac{1}{5}$
• B. $2$
• C. $\in 2$
• D. none of these

Answer 1

Answer: (A)

$-\frac{1}{5}$

$\because y={tan}^{-1}\left(\frac{2^x}{1+2^{2x+1}}\right)\Rightarrow tany=\frac{2^x}{1+2^{2x+1}}⟶\left(1\right)\Rightarrow (1+2^{2x+1})tany=2^{x}differntiatingw.r.tx\Rightarrow {sec}^{2}y(1+2^{2x+1})\frac{dy}{dx}+2(2x+1)2^{2x}tany=x{2}^{x-1}(\frac{d}{dx}a.b=a\frac{db}{dx}+b\frac{da}{dx})\Rightarrow ({tan}^{2}y+1)(1+2^{2x+1})\frac{dy}{dx}+(2x+1)2^{2x+1}tany=x{2}^{x-1}(1+{tan}^{2}x={sec}^{2}x)\Rightarrow (\left(\frac{2^x}{1+2^{2x+1}}\right)+1)(1+2^{2x+1})\frac{dy}{dx}+(2x+1)2^{2x+1}\left(\frac{2^x}{1+2^{2x+1}}\right)=x{2}^{x-1}\left(from\right(1\left)\right)\Rightarrow \frac{dy}{dx}=\frac{x{2}^{x-1}-(2x+1)2^{2x+1}\left(\frac{2^x}{1+2^{2x+1}}\right)}{(\left(\frac{2^x}{1+2^{2x+1}}\right)+1)(1+2^{2x+1})}\Rightarrow {\frac{dy}{dx}}_{x=0}=-\frac{1}{5}$