If x^{2019} cdot y^{2020}=(x+y)^{4039},then f | vedclass

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(C) Given equation: $x^{2019} \cdot y^{2020} = (x+y)^{4039}$.Taking natural logarithm on both sides:$2019 \ln(x) + 2020 \ln(y) = 4039 \ln(x+y)$.Differ

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$\frac{y}{x}$

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(C) Given equation: $x^{2019} \cdot y^{2020} = (x+y)^{4039}$.Taking natural logarithm on both sides:$2019 \ln(x) + 2020 \ln(y) = 4039 \ln(x+y)$.Differentiating both sides with respect to $x$:$\frac{2019}{x} + \frac{2020}{y} \frac{dy}{dx} = 4039 \cdot \frac{1}{x+y} \cdot (1 + \frac{dy}{dx})$.Multiply by $x(x+y)y$ to simplify:$2019(x+y)y + 2020x(x+y) \frac{dy}{dx} = 4039xy(1 + \frac{dy}{dx})$.$2019xy + 2019y^2 + 2020x^2 \frac{dy}{dx} + 2020xy \frac{dy}{dx} = 4039xy + 4039xy \frac{dy}{dx}$.Rearranging terms involving $\frac{dy}{dx}$:$\frac{dy}{dx} (2020x^2 + 2020xy - 4039xy) = 4039xy - 2019xy - 2019y^2$.$\frac{dy}{dx} (2020x^2 - 2019xy) = 2020xy - 2019y^2$.$\frac{dy}{dx} [x(2020x - 2019y)] = y(2020x - 2019y)$.Therefore,$\frac{dy}{dx} = \frac{y}{x}$.

If $x^{2019} \cdot y^{2020}=(x+y)^{4039}$,then $\frac{dy}{dx}=$

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