Differentiation Question 42
Question: If $ y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+…\infty },}} $ then $ \frac{dy}{dx} $ is
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:
[c] $ y=\sqrt{logx+y} $ Or $ y^{2}=\log x+y $ Or $ 2y\frac{dy}{dx}=\frac{1}{x}+\frac{dy}{dx} $ or $ \frac{dy}{dx}=\frac{1}{x(2y-1)} $
BETA
