Limits Continuity And Differentiability Question 48
Question: If $ y={{\tan }^{-1}}( \frac{2^{x}}{1+{2^{2x+1}}} ), $ then $ \frac{dy}{dx}atx=0 $ is
Options:
A) $ \frac{3}{5}\log 2 $
B) $ \frac{2}{5}\log 2 $
C) $ -\frac{3}{2}\log 2 $
D) $ \log 2( \frac{-1}{10} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
Given expression can be written as $ y={{\tan }^{-1}}[ \frac{2^{x}(2-1)}{1+2^{x}{{.2}^{x+1}}} ]={{\tan }^{-1}}[ \frac{{2^{x+1}}-2^{x}}{1+2^{x}{{.2}^{x+1}}} ] $
$ ={{\tan }^{-1}}({2^{x+1}})-ta{n^{-1}}(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 {{( \frac{dy}{dx} )} _{x=0}}=(log2)( \frac{2}{5}-\frac{1}{2} )=\log 2( -\frac{1}{10} ) $
BETA
