Question

If $y^x+x^y+x^x=a^b$, find $\frac{dy}{dx}$.

Answer 1

$$\begin{gathered} \mathrm{Given}\mathrm{that}{y}^x+x^y+x^x=a^b \\ \mathrm{Putting}u=y^x,v=x^y\mathrm{and}w=x^x,\mathrm{we}\mathrm{get} \\ u+v+w=a^b \\ \therefore \frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}=0...\left(i\right) \\ \mathrm{Now},u=y^x \end{gathered}$$
Taking log on both sides,
$\log u=x\log y$

$$\begin{gathered} \Rightarrow \frac{1}{u}\frac{du}{dx}=x\frac{d}{dx}\left(\log y\right)+\log y\frac{d}{dx}\left(x\right)\left[\mathrm{using}\mathrm{product}\mathrm{rule}\right] \\ \Rightarrow \frac{1}{u}\frac{du}{dx}=x\frac{1}{y}\frac{dy}{dx}+\log y\times 1 \\ \Rightarrow \frac{du}{dx}=u\left(\frac{x}{y}\frac{dy}{dx}+\log y\right) \\ \Rightarrow \frac{du}{dx}=y^x\left(\frac{x}{y}\frac{dy}{dx}+\log y\right)...\left(ii\right) \\ \mathrm{Also},v=x^y \end{gathered}$$
Taking log on both sides,
$\log v=y\log x$

$$\begin{gathered} \Rightarrow \frac{1}{v}\frac{dv}{dx}=y\frac{d}{dx}\left(\log x\right)+\log x\frac{dy}{dx} \\ \Rightarrow \frac{1}{v}\frac{dv}{dx}=y\frac{1}{x}+\log x\frac{dy}{dx} \\ \Rightarrow \frac{dv}{dx}=v\left[\frac{y}{x}+\log x\frac{dy}{dx}\right] \\ \Rightarrow \frac{dv}{dx}=x^y\left[\frac{y}{x}+\log x\frac{dy}{dx}\right]...\left(iii\right) \\ \mathrm{Again},w=x^x \end{gathered}$$
Taking log on both sides,
$\log w=x\log x$

$$\begin{gathered} \Rightarrow \frac{1}{w}\frac{dw}{dx}=x\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(x\right) \\ \Rightarrow \frac{1}{w}\frac{dw}{dx}=x\frac{1}{x}+\log x\left(1\right) \\ \Rightarrow \frac{dw}{dx}=w\left(1+\log x\right) \\ \Rightarrow \frac{dw}{dx}=x^x\left(1+\log x\right)...\left(\mathrm{iv}\right) \\ \mathrm{From}\left(\mathrm{i}\right),\left(\mathrm{ii}\right),\left(\mathrm{iii}\right)\mathrm{and}\left(\mathrm{iv}\right),\mathrm{we}\mathrm{have} \\ {y}^x\left(\frac{x}{y}\frac{dy}{dx}+\log y\right)+x^y\left(\frac{y}{x}+\log x\frac{dy}{dx}\right)+x^x\left(1+\log x\right)=0 \\ \Rightarrow \left(x.y^{x-1}+x^y.\log x\right)\frac{dy}{dx}=-x^x\left(1+\log x\right)-y.x^{y-1}-y^x\log y \\ \therefore \frac{dy}{dx}=\frac{-\left\{y^x\log y+y.x^{y-1}+x^x\left(1+\log x\right)\right\}}{x.y^{x-1}+x^y\log x} \end{gathered}$$