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

Question

Aicharya · Accepted Answer

$\log (x+y)=2 x y...(i)$ Differentiating both sides w.r.t. $x$, we get $\begin{aligned} & \left(\frac{1}{x+y}ight)\left(1+\frac{\mathrm{d} y}{\mathrm{~d} x}ight)=2\left(x \frac{\mathrm{d} y}{\mathrm{~d} x}+yight) \ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1-2 x y-2 y^2}{2 x^2+2 x y-1} \end{aligned}$ Putting $x=0$ in (i), we get $\begin{aligned} & y=1 \ 	herefore \quad & \left(\frac{\mathrm{d} y}{\mathrm{~d} x}ight)_{x=0}=\frac{1-0-2}{0+0-1}=1 \end{aligned}$

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

Solution