# If y = $$\sqrt{sinx + \sqrt{sinx + \sqrt{sinx + ……. to \infty}}}$$, find $$dy\over dx$$.

## Solution :

Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as

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

Squaring on both sides,

$$\implies$$  $$y^2$$  = sin x + y

By using differentiation of infinite series,

Differentiating both sides with respect to x,

2y $$dy\over dx$$ =cosx +  $$dy\over dx$$

$$\implies$$ $$dy\over dx$$$$(2y – 1)$$ = cos x

$$\implies$$ $$dy\over dx$$ = $$cos x\over {2y – 1}$$