Question

If $x^x+y^x=1$, prove that $\frac{dy}{dx}=-\left\{\frac{x^x\left(1+\log x\right)+y^x·\log y}{x·y^{\left(x-1\right)}}\right\}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},x^x+y^x=1 \\ \Rightarrow e^{\log x^x}+e^{\log y^x}=1 \\ \Rightarrow e^{x\log x}+e^{x\log y}=1 \end{gathered}$$
Differentiating with respect to x using chain rule,
$$\begin{gathered} \frac{d}{dx}\left(e^{x\log x}\right)+\frac{d}{dx}\left(e^{x\log y}\right)=\frac{d}{dx}\left(1\right) \\ \Rightarrow e^{x\log x}\frac{d}{dx}\left(x\log x\right)+e^{x\log y}\frac{d}{dx}\left(x\log y\right)=0 \\ \Rightarrow e^{x\log x}\left[x\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(x\right)\right]+e^{\log y^x}\left[x\frac{d}{dx}\left(\log y\right)+\log y\frac{d}{dx}\left(x\right)\right]=0 \\ \Rightarrow x^x\left[x\left(\frac{1}{x}\right)+\log x\left(1\right)\right]+y^x\left[x\left(\frac{1}{y}\right)\frac{dy}{dx}+\log y\left(1\right)\right]=0 \\ \Rightarrow x^x\left[1+\log x\right]+y^x\left(\frac{x}{y}\frac{dy}{dx}+\log y\right)=0 \\ \Rightarrow y^x\times \frac{x}{y}\frac{dy}{dx}=-\left[x^x\left(1+\log x\right)+y^x\log y\right] \\ \Rightarrow \left(x{y}^{x-1}\right)\frac{dy}{dx}=-\left[x^x\left(1+\log x\right)+y^x\log y\right] \\ \Rightarrow \frac{dy}{dx}=-\left[\frac{x^x\left(1+\log x\right)+y^x\log y}{x{y}^{x-1}}\right] \end{gathered}$$