Differentiation Question 397
Question: If $ f(x)={{\tan }^{-1}}{ \frac{\log ( \frac{e}{x^{2}} )}{\log (ex^{2})} }+{{\tan }^{-1}}( \frac{3+2\log x}{1-6\log x} ) $ , then $ \frac{d^{n}y}{dx^{n}} $ is $ (n\ge 1) $
Options:
A) $ {{\tan }^{-1}}{{{(\log x)}^{n}}} $
B) 0
C) $ \frac{1}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ y={{\tan }^{-1}}( \frac{\log e-\log x^{2}}{\log e+\log x^{2}} )+{{\tan }^{-1}}( \frac{3+2\log x}{1-6\log x} ) $
$ ={{\tan }^{-1}}( \frac{1-2\log x}{1+2\log x} )+{{\tan }^{-1}}( \frac{3+2\log x}{1-6\log x} ) $
$ ={{\tan }^{-1}}1-{{\tan }^{-1}}(2\log x)+{{\tan }^{-1}}3+{{\tan }^{-1}}(2\log x) $
$ \Rightarrow y={{\tan }^{-1}}1+{{\tan }^{-1}}3\Rightarrow \frac{dy}{dx}=0\Rightarrow \frac{d^{n}y}{dx^{n}}=0. $
BETA
