If $$y^x +x^y+x^x=a^b$$, find $$\dfrac {dy}{dx}$$
We have,
$$y^x+x^y +x^x=a^b$$
$$\Rightarrow \ e^{\log y^x} +e^{\log x^y}+e^{\log x^x}=a^b$$
$$\Rightarrow \ e^{x\log y}+e^{y\log x}+e^{x\log x}=a^b$$
Differentiating both sides with respect to $$x$$, we get
$$\dfrac {d}{dx}(e^{x \log y})+\dfrac {d}{dx} (e^{y \log x})+\dfrac {d}{dx} (e^{x \log x})=\dfrac {d}{dx} a^b)$$
$$\Rightarrow \ e^{x \log y}\dfrac {d}{dx} (x \log y)+e^{y\log x}\dfrac {d}{dx} (y \log x)+e^{x \log x} \dfrac {d}{dx} (x \log x)=0$$
$$\Rightarrow \ y^x \left (\log y+\dfrac {x}{y} \dfrac {dy}{dx} \right) +x^y \left (\dfrac {dy}{dx} \log x+ \dfrac {y}{x} \right) +x^x (1+\log x)=0$$
$$\Rightarrow \ \dfrac {dy}{dx} (xy^{x-1}+x^y \log x)=-\left\{y^x \log y+y\ x^{y-1} +x^x (1+\log x)\right\}$$
$$\Rightarrow \ \dfrac {dy}{dx} =-\dfrac{\left\{y^x \log y+y\ x^{y-1} +x^x (1+\log x)\right\}} {xy^{x-1}+x^y\log x}$$