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

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