If $y$ is a function of $x$ and $\log (x+y)=2 x y$, then the value of $y^{\prime}(0)$ is

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Aicharya · Accepted Answer

$\log (x+y)=2 x y$ Differentiating w.r.t. $x$, we get $\begin{aligned} & \frac{1}{x+y}\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}ight)=2 x \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \ & \frac{1}{x+y}+\frac{1}{(x+y)} \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \ & \left(\frac{1}{x+y}-2 xight) \frac{\mathrm{d} y}{\mathrm{~d} x}=2 y-\frac{1}{x+y} \ & \frac{\mathrm{d} y}{\mathrm{~d} x}\left(\frac{1}{x+y}-2 xight)=2 y-\frac{1}{x+y} \ & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\left(2 y-\frac{1}{x+y}ight)}{\left(\frac{1}{x+y}-2 xight)} \ & 	ext { For } x=0, \log (y)=0 \ & \Rightarrow y=1 \ & \left.\frac{\mathrm{d} y}{\mathrm{~d} x}ight|_{(0,1)}=\frac{\left(2-\frac{1}{0+1}ight)}{\left(\frac{1}{0+1}-0ight)}=1 \end{aligned}$

If $y$ is a function of $x$ and $\log (x+y)=2 x y$, then the value of $y^{\prime}(0)$ is

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