Limits Continuity And Differentiability Question 34
Question: If $ y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+…..\infty }}} $ , then $ \frac{dy}{dx}= $
Options:
A) $ \frac{x}{2y-1} $
B) $ \frac{x}{2y+1} $
C) $ \frac{1}{x(2y-1)} $
D) $ \frac{1}{x(1-2y)} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=\sqrt{\log x+y}\Rightarrow y^{2}=\log x+y $
$ \Rightarrow 2y\frac{dy}{dx}=\frac{1}{x}+\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{1}{x(2y-1)} $
BETA
