(a)OR(b)

If $\displaystyle \mathrm{x}^{30} \mathrm{y}^{20}=(\mathrm{x}+\mathrm{y})^{50}$, prove that $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}$.

Taking log of both sides, we get

$$$30$\displaystyle \log \mathrm{x}+$20$\displaystyle \log \mathrm{y}=$50$ \log (\mathrm{x}+\mathrm{y})

$$Differentiating both sides w.r.t. x , we get

$$\begin{aligned} & \frac{30}{\mathrm{x}}+\frac{20}{\mathrm{y}} \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{50}{\mathrm{x}+\mathrm{y}}\left($\displaystyle 1$+\frac{d \mathrm{y}}{d \mathrm{x}}\right) & \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}\left(\frac{$\displaystyle 20$ \mathrm{x}-$\displaystyle 30$ \mathrm{y}}{\mathrm{y}(\mathrm{x}+\mathrm{y})}\right)=\frac{$\displaystyle 20$ \mathrm{x}-$\displaystyle 30$ \mathrm{y}}{\mathrm{x}(\mathrm{x}+\mathrm{y})} & \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{\mathrm{y}}{\mathrm{x}} \end{aligned}

$$Differentiating both sides w.r.t. x , we get

$$\begin{aligned} & $\displaystyle 5$^{\mathrm{x}} \log $\displaystyle 5$+$\displaystyle 5$^{\mathrm{y}} \log $\displaystyle 5$ \frac{d \mathrm{y}}{d \mathrm{x}}=$\displaystyle 5$^{\mathrm{x}+\mathrm{y}} \log $\displaystyle 5$\left($\displaystyle 1$+\frac{d \mathrm{y}}{d \mathrm{x}}\right) & \Rightarrow $\displaystyle 5$^{\mathrm{x}}+$\displaystyle 5$^{\mathrm{y}} \frac{d \mathrm{y}}{d \mathrm{x}}=\left($\displaystyle 5$^{\mathrm{x}}+$\displaystyle 5$^{\mathrm{y}}\right)\left($\displaystyle 1$+\frac{d \mathrm{y}}{d \mathrm{x}}\right) & \Rightarrow $\displaystyle 5$^{\mathrm{x}}+$\displaystyle 5$^{\mathrm{y}} \frac{d \mathrm{y}}{d \mathrm{x}}=$\displaystyle 5$^{\mathrm{x}}+$\displaystyle 5$^{\mathrm{x}} \frac{d \mathrm{y}}{d \mathrm{x}}+$\displaystyle 5$^{\mathrm{y}}+$\displaystyle 5$^{\mathrm{y}} \frac{d \mathrm{y}}{d \mathrm{x}} & \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=-\frac{$\displaystyle 5$^{\mathrm{y}}}{$\displaystyle 5$^{\mathrm{x}}}=-$\displaystyle 5$^{\mathrm{y}-\mathrm{x}} \end{aligned}