Practicing Success
Practicing Success
CUET Preparation Today
Let $y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+.....}}}$, then $\frac{dy}{dx}$ is: |
$\frac{2y}{\cos x}$ $\frac{\cos x}{2y-1}$ $\frac{2y-1}{2y+1}$ $\frac{\cos x}{2y+1}$ |
$\frac{\cos x}{2y-1}$ |
$y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+.....}}}=\sqrt{\sin x+y}$ $⇒y^2=\sin x+y⇒2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}$ $⇒\frac{dy}{dx}=\frac{\cos x}{2y-1}$ |
