Differentiation of Infinite Series

Calculus Level 2

$y=\sqrt { \sin x+\sqrt { \sin x+\sqrt { \sin x + \sqrt{\cdots} } } }$

What is the derivative (with respect to $x$ ) of the above infinitely nested expression?

\frac { { d }^{ \infty }y }{ { { dx }^{ \infty } } }

\sqrt { \frac { \cos x }{ \ln { \cos x } } }

\sqrt { \sin x }

\frac { \cos x }{ 2y-1 }

1 solution

Md Zuhair
Jan 12, 2017

This problem has very easy solution.

Here $y=\sqrt { \sin x+\sqrt { \sin x+\sqrt { \sin x + \sqrt{\cdots} } } }$

Now , $y=\sqrt { \sin x+y}$ or $y^2 = sinx + y$

Differentiating both sides with respect to x we get,

$2y\dfrac{dy}{dx} = cosx + \dfrac{dy}{dx}$

Hence after solving for $\dfrac{dy}{dx}$ we get

$\dfrac{dy}{dx}$ = $\dfrac{cosx}{2y-1}$ (Ans)