If $\log (x+y)-2 x y=0$, then $y^{\prime}(0)=$

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Aicharya · Accepted Answer

$\log (x+y)-2 x y=0$ Differentiating with respect to ' $x$ ' we get, $\begin{aligned} \frac{1}{x+y}\left(1+\frac{d y}{d x}\right)-2\left(x \frac{d y}{d x}+y\right) & =0 \ \Rightarrow \quad \frac{1}{x+y}+\left(\frac{1}{x+y}-2 x\right) \frac{d y}{d x}-2 y & =0 \end{aligned}$ $\begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=\frac{2 y-\frac{1}{x+y}}{\frac{1}{x+y}-2 x} \ & \left.\Rightarrow \quad \frac{d y}{d x}\right|_{x=0}=\frac{2 y-\frac{1}{y}}{\frac{1}{y}}=2 y^2-1\end{aligned}$

If \(\log (x+y)-2 x y=0\), then \(y^{\prime}(0)=\)

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