If $\log (x+y)=2 x y$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=0$ is

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Aicharya · Accepted Answer

$\log (x+y)=2 x y$ Differentiating w.r.t. $x$, we get $\begin{array}{ll} & \frac{1}{x+y}\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}ight)=2 y+2 x \frac{\mathrm{d} y}{\mathrm{~d} x} \ 	herefore \quad & \frac{1}{x+y}+\frac{1}{x+y} \frac{\mathrm{d} y}{\mathrm{~d} x}=2 y+2 x \frac{\mathrm{d} y}{\mathrm{~d} x} \ & 	ext { At } x=0,(\mathrm{i}) \Rightarrow y=1 \ 	herefore \quad & 	ext { (ii) }\left.\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}ight|_{x=0}=1 \end{array}$

If $\log (x+y)=2 x y$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=0$ is

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