Skip to main contentSkip to solution

If y=sin⁡x+sin⁡x+sin⁡x+⋯+∞y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots + \infty}}} then dydx\frac{dy}{dx} equals to

Solution

✅ Correct Option: 2

Given y=sin⁡x+sin⁡x+sin⁡x+⋯+∞y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots + \infty}}}

The expression inside the first square root contains the same infinite pattern as the original expression:

y=sin⁡x+sin⁡x+sin⁡x+⋯⏟equals yy = \sqrt{\sin x + \underbrace{\sqrt{\sin x + \sqrt{\sin x + \cdots}}}_{\text{equals } y}}

Therefore:

y=sin⁡x+yy = \sqrt{\sin x + y}


y=sin⁡x+yy = \sqrt{\sin x + y}

y2=sin⁡x+yy^2 = \sin x + y


Differentiating both sides with respect to xx:

ddx(y2)=ddx(sin⁡x+y)\frac{d}{dx}(y^2) = \frac{d}{dx}(\sin x + y)

2ydydx=cos⁡x+dydx2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}


2ydydx=cos⁡x+dydx2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}

2ydydx−dydx=cos⁡x2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x

dydx(2y−1)=cos⁡x\frac{dy}{dx}(2y - 1) = \cos x

dydx=cos⁡x2y−1\frac{dy}{dx} = \frac{\cos x}{2y - 1}

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question