Question

यदि y = /sinx+y , तो y =​

Answer 1

Answer:

$y = \sqrt{sinx + y} \\ \therefore \: y \: = {(sinx + y)}^{ \frac{1}{2} } \\ differentiating \: w.r.t.x \: on \: both \: \\ sides. \\ \frac{dy}{dx} = \frac{d}{dx} {(sinx + y)}^{ \frac{1}{2} } \\ = \frac{1}{2} {(sinx + y)}^{ - \frac{ 1}{2} } \frac{d}{dx} {(sinx + y)} \\ = \frac{1}{{2(sinx + y)}^{ \frac{ 1}{2} }} (cosx + \frac{dy}{dx} ) \\ = \frac{1}{2 \sqrt{sinx + y} } (cosx + \frac{dy}{dx} ) \: \\ 2 \sqrt{sinx + y} \frac{dy}{dx} = cosx + \frac{dy}{dx} \\ 2 \sqrt{sinx + y} \frac{dy}{dx} - \frac{dy}{dx} = cosx \\ (2 \sqrt{sinx + y} - 1)\frac{dy}{dx} = cosx \\ \frac{dy}{dx} = \frac{cosx}{2 \sqrt{sinx + y} - 1}$