If \(\rm y = \sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+...\infty}}}\), then \(\rm \dfrac{dy}{dx}=\)

Question

Question

\(\rm \dfrac{\cos x}{y-1}\)

duplicate options found. English Question 1 options 2,4
\(\rm \dfrac{\cos x}{2y-1}\)
\(\rm \dfrac{\sin x}{2y-1}\)
\(\rm \dfrac{\cos x}{2y+1}\)

Answer 1

Concept:

\(\rm \dfrac{d(\sin x)}{dx}= \cos x\)

Calculations:

Given, \(\rm y = \sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+...\infty}}}\)

⇒ \(\rm y = \sqrt{\sin x+y}\)

Squaring both side in above the equation,

⇒ \(\rm y^2 =\sin x+y\)

⇒ \(\rm 2y\dfrac{dy}{dx} = \cos x + \dfrac{dy}{dx}\)

⇒ \(\rm (2y-1)\dfrac{dy}{dx} = \cos x\)

⇒ \(\rm \dfrac{dy}{dx} = \dfrac {\cos x}{2y-1} \)

Hence, if \(\rm y = \sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+...\infty}}}\) then \(\rm \dfrac{dy}{dx}=\)\(\rm \dfrac{\cos x}{2y-1}\)