A) \[\frac{x}{2y-1}\]
B) \[\frac{1}{2y-1}\]
C) \[\frac{2}{y-1}\]
D) None of these
Correct Answer: B
Solution :
\[y=\sqrt{x+\sqrt{x+\sqrt{x+.....\infty }}}\] \[{{y}^{2}}=x+\sqrt{x+\sqrt{x+.....\infty }}\] \[\Rightarrow \] \[{{y}^{2}}=x+y\] On differentiating with respect to\[x,\] \[2y\frac{dy}{dx}=1+\frac{dy}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}(2y-1)=1\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{2y-1}\]
You need to login to perform this action.
You will be redirected in
3 sec
