Applications Of Derivatives Question 200
Question: If $ y={{(x\log x)}^{\log ,\log x}} $ , then $ \frac{dy}{dx}= $
[Roorkee 1981]
Options:
A) $ {{(x\log x)}^{\log \log x}}{ \frac{1}{x\log x}(\log x+\log \log x)+(\log \log x)( \frac{1}{x}+\frac{1}{x\log x} ) } $
B) $ {{(x\log x)}^{x\log x}}\log \log x[ \frac{2}{\log x}+\frac{1}{x} ] $
C) $ {{(x\log x)}^{x\log x}}\log \log x[ \frac{2}{\log x}+\frac{1}{x} ] $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={{(x\log x)}^{\log \log x}} $
therefore $ \log y=\log \log x[\log x+\log \log x] $
therefore $ \frac{1}{y}\frac{dy}{dx}=\frac{1}{x\log x}(\log x+\log \log x)+\log \log x( \frac{1}{x}+\frac{1}{x\log x} ) $
therefore $ s^{2}=at^{2}+bt+c $ .
BETA
