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