Differentiation Question 253
Question: If $ y=\frac{e^{2x}\cos x}{x\sin x}, $ then $ \frac{dy}{dx}= $
[AI CBSE 1982]
Options:
A) $ \frac{e^{2x}[(2x-1)\cot x-xcose{c^{2}}x]}{x^{2}} $
B) $ \frac{e^{2x}[(2x+1)\cot x-xcose{c^{2}}x]}{x^{2}} $
C) $ \frac{e^{2x}[(2x-1)\cot x+xcose{c^{2}}x]}{x^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y=\frac{e^{2x}\cos x}{x\sin x} $
Therefore $ \log y=2x+\log \cos x-\log x-\log \sin x $
$ \frac{1}{y}\frac{dy}{dx}=2+( \frac{-\sin x}{\cos x} )-\frac{1}{x}-\frac{\cos x}{\sin x} $
Therefore $ \frac{dy}{dx}=e^{2x}[ \frac{2}{x}\cot x-\frac{1}{x}-\frac{1}{x^{2}}\cot x-\frac{{{\cot }^{2}}x}{x} ] $
$ =\frac{e^{2x}}{x^{2}}[(2x-1)\cot x-xcose{c^{2}}x] $ .
BETA
