If log (x+y)=2xy,then frac{dy}{dx} at x=0 is | vedclass

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(A) Given equation: $\log(x+y) = 2xy$ $(i)$ Differentiating both sides with respect to $x$: $\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = 2y + 2x \f

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$1$

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(A) Given equation: $\log(x+y) = 2xy$ $(i)$ Differentiating both sides with respect to $x$: $\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = 2y + 2x \frac{dy}{dx}$ $(ii)$ At $x=0$,from $(i)$: $\log(0+y) = 2(0)y \implies \log(y) = 0 \implies y = e^0 = 1$. Substituting $x=0$ and $y=1$ into $(ii)$: $\frac{1}{0+1} \left(1 + \frac{dy}{dx}\right) = 2(1) + 2(0) \frac{dy}{dx}$ $1 + \frac{dy}{dx} = 2$ $\frac{dy}{dx} = 2 - 1 = 1$ Thus,the value of $\frac{dy}{dx}$ at $x=0$ is $1$.

If $\log (x+y)=2xy$,then $\frac{dy}{dx}$ at $x=0$ is

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