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If $y$ is a function of $x$ and $\log (x+y)=2 x y$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=0$ is
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$1$
$\log (x+y)=2 x y...(i)$
Differentiating both sides w.r.t. $x$, we get
$\begin{aligned}
& \left(\frac{1}{x+y}\right)\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2\left(x \frac{\mathrm{d} y}{\mathrm{~d} x}+y\right) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1-2 x y-2 y^2}{2 x^2+2 x y-1}
\end{aligned}$
Putting $x=0$ in (i), we get
$\begin{aligned}
& y=1 \\
\therefore \quad & \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=\frac{1-0-2}{0+0-1}=1
\end{aligned}$
Differentiating both sides w.r.t. $x$, we get
$\begin{aligned}
& \left(\frac{1}{x+y}\right)\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2\left(x \frac{\mathrm{d} y}{\mathrm{~d} x}+y\right) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1-2 x y-2 y^2}{2 x^2+2 x y-1}
\end{aligned}$
Putting $x=0$ in (i), we get
$\begin{aligned}
& y=1 \\
\therefore \quad & \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=0}=\frac{1-0-2}{0+0-1}=1
\end{aligned}$
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