Differentiation Question 433
Question: If $ y={e^{x+{e^{x+{e^{x+….\infty }}}}}} $ , then $ \frac{dy}{dx}= $
[AISSE 1990; UPSEAT 2002; DCE 2002]
Options:
A) $ \frac{y}{1-y} $
B) $ \frac{1}{1-y} $
C) $ \frac{y}{1+y} $
D) $ \frac{y}{y-1} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={e^{x+y}} $
Therefore $ \log y=(x+y)\log e $
Therefore $ \frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{dx} $
Therefore $ ={{\sin }^{2}}\alpha +\frac{1}{2}(\cos 2\alpha +\cos 2\beta ) $ .
BETA
