Differentiation Question 20
Question: If $ y=\sqrt{x+\sqrt{x+\sqrt{x+……..to}}}\infty ,then\frac{dy}{dx}= $
[RPET 2002]
Options:
A) $ \frac{x}{2y-1} $
B) $ \frac{2}{2y-1} $
C) $ \frac{-1}{2y-1} $
D) $ \frac{1}{2y-1} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ y=\sqrt{x+\sqrt{x+\sqrt{x+……\infty }}} $
Therefore $ y=\sqrt{x+y} $
Therefore $ \frac{{{\partial }^{2}}u}{\partial y^{2}}=\frac{1}{u}-\frac{{{(y-b)}^{2}}}{u^{3}} $
Therefore $ 2y\frac{dy}{dx}=1+\frac{dy}{dx} $
Therefore $ \frac{dy}{dx}(2y-1)=1 $
Therefore $ \frac{dy}{dx}=\frac{1}{2y-1} $ .
BETA
