Differentiation Question 22
Question: If $ x={e^{y+{e^{y+….to\infty }}}} $ , $ x>0, $ then $ \frac{dy}{dx} $ is
[AIEEE 2004]
Options:
A) $ \frac{1+x}{x} $
B) $ \frac{1}{x} $
C) $ \frac{1-x}{x} $
D) $ \frac{x}{1+x} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ x={e^{y+{e^{y+….to\infty }}}} $ , $ x>0 $ , $ x={e^{y+x}} $
Taking log to the both sides, $ \log x=(y+x) $
Differentiate both sides w.r.t. x, $ \frac{1}{x}=\frac{dy}{dx}+1 $
$ \Rightarrow \frac{dy}{dx}=\frac{1-x}{x} $ .
BETA
