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

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Aicharya · Accepted Answer

$x^{2019} \cdot y^{2020}=(x+y)^{4039}$ On differentiating $\begin{gathered} \Rightarrow 2020 y^{2019} \frac{d y}{d x} x^{2019}+2019 x^{2018} \cdot y^{2020} \ =4039(x+y)^{4038} \cdot\left(1+\frac{d y}{d x}\right) \ \Rightarrow \quad 2020 \frac{(x+y)^{4039}}{y} \frac{d y}{d x}+2019 \frac{(x+y)^{4039}}{x} \ =4039(x+y)^{4038}\left(1+\frac{d y}{d x}\right) \ \Rightarrow \quad\left(\frac{2020}{y} \frac{d y}{d x}+\frac{2019}{x}\right)(x+y)=4039\left(1+\frac{d y}{d x}\right) \ \Rightarrow \quad \frac{2020(x+y)}{y} \cdot \frac{d y}{d x}+\frac{2019}{x}(x+y)=4039\left(1+\frac{d y}{d x}\right) \ \Rightarrow \quad \frac{d y}{d x}\left[4039-\frac{2020(x+y)}{y}\right]=\frac{2019(x+y)}{x} \ \Rightarrow \frac{d y}{d x}\left[\frac{2019 y-2020 x}{y}\right]=\frac{2019(x+y)}{x}-4039 \ \Rightarrow \frac{d y}{d x}=\frac{y}{x} \end{gathered}$

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

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