If y is a function of x and log(x+y)=2xy,then | vedclass

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(A) Given equation: $\log(x+y)=2xy$ Differentiating both sides with respect to $x$: $\frac{1}{x+y} \cdot (1 + y') = 2y + 2xy'$ At $x=0$,the equation b

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$1$

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(A) Given equation: $\log(x+y)=2xy$ Differentiating both sides with respect to $x$: $\frac{1}{x+y} \cdot (1 + y') = 2y + 2xy'$ At $x=0$,the equation becomes $\log(0+y) = 2(0)y$,which implies $\log(y) = 0$,so $y = e^0 = 1$. Substituting $x=0$ and $y=1$ into the differentiated equation: $\frac{1}{0+1} \cdot (1 + y'(0)) = 2(1) + 2(0)y'(0)$ $1 \cdot (1 + y'(0)) = 2 + 0$ $1 + y'(0) = 2$ $y'(0) = 2 - 1 = 1$

If $y$ is a function of $x$ and $\log(x+y)=2xy$,then the value of $y'(0)$ is

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