Question

If $y=\sqrt{\sin x-\sqrt{\sin x-\sqrt{\sin x-\cdots \text{to}\infty }}}$ , then $\frac{dy}{dx}=$

• A. $\frac{\sin x}{2y+1}$
• B. $\frac{\cos x}{2y+1}$
• C. $\frac{{\cos }^{2}x}{2y-1}$
• D. $\frac{{\sin }^{2}x}{2y-1}$

Answer 1

Answer: (B)

$\frac{\cos x}{2y+1}$

The given information is:

$y=\sqrt{\sin x-\sqrt{\sin x-\sqrt{\sin x-...\infty }}}$

As the function keeps on repeating to infinity we can write the terms leaving the first term as y and it would not make any difference so we have,

$\Rightarrow y=\sqrt{\sin x-y}$

$\Rightarrow y^2=\sin x-y$

$\Rightarrow y^2+y=\sin x$

Differentiating with respect to $x$ we get,

$\Rightarrow (2y+1)\frac{dy}{dx}=\cos x$

$\Rightarrow \frac{dy}{dx}=\frac{\cos x}{(2y+1)}$ .....Answer