A) \[\frac{x}{2y-1}\]
B) \[\frac{2}{2y-1}\]
C) \[\frac{-1}{2y-1}\]
D) \[\frac{1}{2y-1}\]
Correct Answer: D
Solution :
\[y=\sqrt{x+\sqrt{x+\sqrt{x+......\infty }}}\] Þ \[y=\sqrt{x+y}\] Þ \[\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=\frac{1}{u}-\frac{{{(y-b)}^{2}}}{{{u}^{3}}}\] Þ \[2y\frac{dy}{dx}=1+\frac{dy}{dx}\] Þ \[\frac{dy}{dx}(2y-1)=1\] Þ \[\frac{dy}{dx}=\frac{1}{2y-1}\].
You need to login to perform this action.
You will be redirected in
3 sec
