If x^y-y^x=a^b, find = dy dx .

Question

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

✓

Answer

The given function is $\text{x}^\text{y}-\text{y}^x=\text{a}^\text{b}$
Let $\text{x}^\text{y}=\text{u}$ and $\text{y}^\text{x}=\text{v}$
Then, the function becomes $\text{u}-\text{v}=\text{a}^\text{b}$
$\frac{\text{du}}{\text{dx}}-\frac{\text{dv}}{\text{dx}}=0\ \dots(1)$
$\text{u}=\text{x}^\text{y}$
$\Rightarrow\log\text{u}=\log(\text{x}^\text{y})$
$\Rightarrow\log\text{u}=\text{y}\log\text{x}$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{u}}.\frac{\text{du}}{\text{dx}}=\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}.\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}.\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^\text{y}\Big(\log\text{x}\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}\Big)\ \dots(2)$
$\text{v}=\text{y}^\text{x}$
$\Rightarrow\log\text{v}=\log(\text{y}^\text{x})$
$\Rightarrow\log\text{v}=\text{x}\log\text{y}$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}.\frac{\text{dv}}{\text{dx}}=\log\text{y}.\frac{\text{dy}}{\text{dx}}(\text{x})+\text{x}.\frac{\text{d}}{\text{dx}}(\log\text{y})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big(\log\text{y}.1+\text{x}.\frac{1}{\text{y}}.\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{y}^\text{x}\Big(\log\text{y}+\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}\Big)\ \dots(3)$
From (1), (2) and (3), we obtain
$\text{x}^\text{y}\Big(\log\text{x}\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}\Big)-\text{y}^\text{x}\Big(\log\text{y}+\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow\text{x}^\text{y}\log\text{x}\frac{\text{dy}}{\text{dx}}-\text{x}\text{y}^{\text{x}-1}\frac{\text{dy}}{\text{dx}}+\text{x}^{\text{y}-1}\text{y}-\text{y}^\text{x}\log\text{y}=0$
$\Rightarrow(\text{x}^\text{y}\log\text{x}-\text{xy}^{\text{x}-1})\frac{\text{dy}}{\text{dx}}=\text{y}^\text{x}\log\text{y}-\text{x}^{\text{y}-1}\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^\text{x}\log\text{y}-\text{x}^{\text{y}-1}\text{y}}{(\text{x}^\text{y}\log\text{x}-\text{xy}^{\text{x}-1})}$