Question
If $$y = \sqrt {\log \,x + \sqrt {\log \,x + \sqrt {\log \,x + .....\infty } } } ,$$ then $$\frac{{dy}}{{dx}} = ?$$
A.
$$\frac{x}{{2y - 1}}$$
B.
$$\frac{x}{{2y + 1}}$$
C.
$$\frac{1}{{x\left( {2y - 1} \right)}}$$
D.
$$\frac{1}{{x\left( {1 - 2y} \right)}}$$
Answer :
$$\frac{1}{{x\left( {2y - 1} \right)}}$$
Solution :
$$\eqalign{
& y = \sqrt {\log \,x + y} \cr
& \Rightarrow {y^2} = \log \,x + y \cr
& \Rightarrow 2y\frac{{dy}}{{dx}} = \frac{1}{x} + \frac{{dy}}{{dx}} \cr
& \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{x\left( {2y - 1} \right)}} \cr} $$