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

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(C) Given equation: $\log(x+y)=2xy$ ... $(i)$ At $x=0$,$\log(0+y)=2(0)y$,which implies $\log(y)=0$,so $y=e^0=1$. Differentiating both sides of $(i)$ w

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$1$

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

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

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