If y = √(sin x + y), then dy/dx = A. sin x/2y - 1 B. sin x/(1-2y) C. (cos x)/(1 - 2y) D. (cos x)/(2y - 1)

Question

If y = \(\sqrt{sin\,\text x+y,}\), then \(\cfrac{dy}{d\mathrm x} \) =

A.\(\cfrac{sin\,\text x}{2y-1}\)

B. \(\cfrac{sin\,\text x}{1-2y}\)

C. \(\cfrac{cos\,\text x}{1-2y}\)

D. \(\cfrac{cos\,\text x}{2y-1}\)

Answer 1

 Correct option is D.\(\cfrac{cos\,\text x}{2y-1}\)

 y = \(\sqrt{sin\,\text x+y}\)

Squaring both sides

⇒ y² = sin x + y

Differentiating w.r.t x we get,