A) \[-\frac{3}{5}\log 2\]
B) \[\frac{2}{5}\log 2\]
C) \[-\frac{3}{2}\log 2\]
D) None
Correct Answer: A
Solution :
\[y={{\tan }^{-1}}\left[ \frac{{{2}^{x}}(2-1)}{1+{{2}^{x}}{{.2}^{x+1}}} \right]={{\tan }^{-1}}\left[ \frac{{{2}^{x+1}}-{{2}^{x}}}{1+{{2}^{x}}{{.2}^{x+1}}} \right]\] \[={{\tan }^{-1}}({{2}^{x+1}})-{{\tan }^{-2}}({{2}^{x}})\Rightarrow \frac{dy}{dx}=\frac{{{2}^{x+1}}\log 2}{1+{{2}^{2(x+1)}}}-\frac{{{2}^{x}}\log 2}{1+{{2}^{2x}}}\]\[\therefore \]\[{{\left( \frac{dy}{dx} \right)}_{x=0}}=(\log 2)\left( \frac{2}{5}-1 \right)=\log 2\left( -\frac{3}{5} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
