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Question Number 129595 by Adel last updated on 16/Jan/21

$$\mathrm{y}=\sqrt{\sin \mathrm{x}+\sqrt{\sin \mathrm{x}+\sqrt{\sin \mathrm{x}+.............\mathrm{\infty }}}}$$$$$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=?$$

Answered by ajfour last updated on 16/Jan/21

$$y^2-\sin x=y$$$$2y\frac{dy}{dx}-\cos x=\frac{dy}{dx}$$$$\frac{dy}{dx}=\frac{\cos x}{2y-1}$$

Answered by mr W last updated on 16/Jan/21

$$y^2=\sin x+y$$$$y^2-y-\sin x=0$$$$y=\frac{1+\sqrt{1+4\sin x}}{2}$$$$\Rightarrow \frac{dy}{dx}=\frac{\cos x}{\sqrt{1+4\sin x}}$$

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