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

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(A) Given the equation: $\log (x+y)=2xy \quad ...(i)$Differentiating both sides with respect to $x$ using the chain rule and product rule:$\frac{1}{x+

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(A) Given the equation: $\log (x+y)=2xy \quad ...(i)$Differentiating both sides with respect to $x$ using the chain rule and product rule:$\frac{1}{x+y} \cdot (1 + y^{\prime}) = 2(x y^{\prime} + y)$Rearranging to solve for $y^{\prime}$:$1 + y^{\prime} = 2(x+y)(x y^{\prime} + y)$$1 + y^{\prime} = 2x^2 y^{\prime} + 2xy + 2xy y^{\prime} + 2y^2$$y^{\prime}(1 - 2x^2 - 2xy) = 2xy + 2y^2 - 1$$y^{\prime} = \frac{2xy + 2y^2 - 1}{1 - 2x^2 - 2xy}$Now,find the value of $y$ when $x=0$ from equation $(i)$:$\log(0+y) = 2(0)y$$\log(y) = 0$$y = e^0 = 1$Substitute $x=0$ and $y=1$ into the expression for $y^{\prime}$:$y^{\prime}(0) = \frac{2(0)(1) + 2(1)^2 - 1}{1 - 2(0)^2 - 2(0)(1)}$$y^{\prime}(0) = \frac{0 + 2 - 1}{1 - 0 - 0} = \frac{1}{1} = 1$

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

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