If ln (x + y) = 2xy,then y'(0) = | vedclass

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(A) Given the equation $\ln (x + y) = 2xy$. First,find the value of $y$ when $x = 0$: $\ln (0 + y) = 2(0)y \implies \ln (y) = 0 \implies y = e^0 = 1$.

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$1$

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(A) Given the equation $\ln (x + y) = 2xy$. First,find the value of $y$ when $x = 0$: $\ln (0 + y) = 2(0)y \implies \ln (y) = 0 \implies y = e^0 = 1$. Now,differentiate both sides with respect to $x$: $\frac{d}{dx} [\ln (x + y)] = \frac{d}{dx} [2xy]$ $\frac{1}{x + y} \cdot (1 + y') = 2(y + xy')$ Substitute $x = 0$ and $y = 1$ into the differentiated equation: $\frac{1}{0 + 1} \cdot (1 + y'(0)) = 2(1 + 0 \cdot y'(0))$ $1 \cdot (1 + y'(0)) = 2(1)$ $1 + y'(0) = 2$ $y'(0) = 2 - 1 = 1$.

If $\ln (x + y) = 2xy$,then $y'(0) =$

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