If $\log (x+y)=\sin (x+y)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ is

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Aicharya · Accepted Answer

$\log (x+y)=\sin (x+y)$
Differentiating both sides w.r.t: $x$, we get $\begin{aligned} & \frac{1}{x+y}\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=\cos (x+y)\left[1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right] \\ & \Rightarrow \frac{1}{x+y}+\frac{1}{x+y} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}=\cos (x+y)+\cos (x+y) \frac{\mathrm{d} y}{\mathrm{~d} x} \\ & \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}\left(\frac{1}{x+y}-\cos (x+y)\right)=\cos (x+y)-\frac{1}{x+y} \\ & \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=-1 \end{aligned}$

If $\log (x+y)=\sin (x+y)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ is

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