If $y=\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{\cos x}}}$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is…
If $y=\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{\cos x}}}$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is given by
- $\frac{y \sin x+(1+y) \cos x}{1+2 y+\cos x-\sin x}$
- $\frac{y \cos x+(1+y) \sin x}{1+2 y+\cos x-\sin x}$
- $\frac{y \sin x-(1+y) \cos x}{1+2 y+\cos x-\sin x}$
- $\frac{y \cos x-(1+y) \sin x}{1+2 y+\cos x-\sin x}$
Solution
$\begin{aligned} & y=\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{1+\frac{\cos x}{\ldots \ldots}}}} \\ & y=\frac{\sin x}{1+\frac{\cos x}{1+y}}\end{aligned}$
$\begin{aligned}
& y=\frac{(1+y) \sin x}{(1+y)+\cos x} \\
& \dot{y}((1+y)+\cos x)=(1+y) \sin x \\
& y+y^2+y \cos x=\sin x+\sin x \cdot y
\end{aligned}$
Differentiating w.r.to $x$, we get
$\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}+y(-\sin x)+\cos x \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}$
$=\frac{\mathrm{d} y}{\mathrm{~d} x} \sin x+(1+y) \cos x$
$\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}-y \sin x+\cos x \frac{\mathrm{~d} y}{\mathrm{~d} x} \\ & =\sin x \frac{\mathrm{~d} y}{\mathrm{~d} x}+(1+y) \cos x \\ & \frac{\mathrm{~d} y}{\mathrm{~d} x}(1+2 y+\cos x-\sin x)\end{aligned}$
$=\cos x+y \cos x+y \sin x$
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y \sin x+(1+y) \cos x}{1+2 y+\cos x-\sin x}$
Asked in: MHT CET 2024 (10 May Shift 1)
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