If $\quad y=\sqrt{(x-\sin x)+\sqrt{(x-\sin x)+\sqrt{(x-\sin x) \ldots .}}}$, then $\frac{\mathrm{d}…
If $\quad y=\sqrt{(x-\sin x)+\sqrt{(x-\sin x)+\sqrt{(x-\sin x) \ldots .}}}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$
- $\frac{1-\cos x}{2 y-1}$
- $\frac{1+\cos x}{2 y-1}$
- $\frac{1-\cos x}{2 y+1}$
- $\frac{1-\sin x}{2 y-1}$
Solution
$\begin{aligned} & \quad \text { If } y=\sqrt{\mathrm{f}(x)+\sqrt{\mathrm{f}(x)+\sqrt{\mathrm{f}(x)+\ldots}}}, \text { then } \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{f}^{\prime}(x)}{2 y-1} \\ & \quad y=\sqrt{(x-\sin x)+\sqrt{(x-\sin x)+\sqrt{(x-\sin x)+\ldots}}} \\ & \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1-\cos x}{2 y-1}\end{aligned}$
Asked in: MHT CET 2023 (14 May Shift 2)
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