Differentiation Question 251
Question: If $ y=\frac{e^{x}\log x}{x^{2}} $ , then $ \frac{dy}{dx}= $
[AI CBSE 1982]
Options:
A) $ \frac{e^{x}[1+(x+2)\log x]}{x^{3}} $
B) $ \frac{e^{x}[1-(x-2)\log x]}{x^{4}} $
C) $ \frac{e^{x}[1-(x-2)\log x]}{x^{3}} $
D) $ \frac{e^{x}[1+(x-2)\log x]}{x^{3}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{dy}{dx}=-2{x^{-3}}e^{x}\log x+{x^{-2}}( e^{x}\log x+\frac{e^{x}}{x} ) $
$ =e^{x}[ \frac{1+(x-2)\log x}{x^{3}} ] $
Aliter: Taking $ \log $ , $ \log y=x+\log \log x-2\log x $
Therefore $ \frac{1}{y}\frac{dy}{dx}=1+\frac{1}{x\log x}-\frac{2}{x} $
Therefore $ \frac{dy}{dx}=\frac{e^{x}\log x}{x^{2}} $ , $ [ \frac{x\log x+1-2\log x}{x\log x} ] $
$ =\frac{e^{x}[(x-2)\log x+1]}{x^{3}} $ .
BETA
