Given that $\displaystyle x^{\mathrm{y}}+\mathrm{y}^{x}=a^{b}$, where $\displaystyle a$ and $\displaystyle b$ are positive constants, find $\displaystyle \frac{d \mathrm{y}}{d x}$.

Let $\displaystyle \mathrm{u}=x^{\mathrm{y}} \Rightarrow \frac{d \mathrm{u}}{d x}=x^{\mathrm{y}}\left(\frac{\mathrm{y}}{x}+\log x \cdot \frac{d \mathrm{y}}{d x}\right), \mathrm{v}=\mathrm{y}^{x} \Rightarrow \frac{d \mathrm{v}}{d x}=\mathrm{y}^{x}\left(\frac{x}{\mathrm{y}} \frac{d \mathrm{y}}{d x}+\log \mathrm{y}\right)$ Since, $\displaystyle \mathrm{u}+\mathrm{v}=a^{b} \Rightarrow \frac{d \mathrm{u}}{d x}+\frac{d \mathrm{v}}{d x}=0$ $\displaystyle \therefore \frac{d \mathrm{y}}{d x}=\frac{-\left(x^{\mathrm{y}-1} \mathrm{y}+\mathrm{y}^{x} \log \mathrm{y}\right)}{\left(x^{\mathrm{y}} \log x+\mathrm{y}^{x-1} x\right)}$