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If \(\log (x+y)-2 x y=0\), then \(y^{\prime}(0)=\)
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Verified Answer
The correct answer is:
\(2 y^2-1\)
\(\log (x+y)-2 x y=0\)
Differentiating with respect to ' \(x\) ' we get,
\(\begin{aligned}
\frac{1}{x+y}\left(1+\frac{d y}{d x}\right)-2\left(x \frac{d y}{d x}+y\right) & =0 \\
\Rightarrow \quad \frac{1}{x+y}+\left(\frac{1}{x+y}-2 x\right) \frac{d y}{d x}-2 y & =0
\end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=\frac{2 y-\frac{1}{x+y}}{\frac{1}{x+y}-2 x} \\ & \left.\Rightarrow \quad \frac{d y}{d x}\right|_{x=0}=\frac{2 y-\frac{1}{y}}{\frac{1}{y}}=2 y^2-1\end{aligned}\)
Differentiating with respect to ' \(x\) ' we get,
\(\begin{aligned}
\frac{1}{x+y}\left(1+\frac{d y}{d x}\right)-2\left(x \frac{d y}{d x}+y\right) & =0 \\
\Rightarrow \quad \frac{1}{x+y}+\left(\frac{1}{x+y}-2 x\right) \frac{d y}{d x}-2 y & =0
\end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=\frac{2 y-\frac{1}{x+y}}{\frac{1}{x+y}-2 x} \\ & \left.\Rightarrow \quad \frac{d y}{d x}\right|_{x=0}=\frac{2 y-\frac{1}{y}}{\frac{1}{y}}=2 y^2-1\end{aligned}\)
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