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

Question

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

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Answer

$x^y + y^x = ab$
Let $u + v = a^b$, where $x^y = u$ and $y^x = v.$
$\therefore \frac{\text{du}}{\text{dx}} + \frac{\text{dv}}{\text{dx}} = 0 \text{ }\text{ }\text{ }\text{ }\text{ }\dots\text{(i)}$
$\text{y} = \log \text{x} = \log \text{u} \Rightarrow \frac{\text{du}}{\text{dx}}=\text{x}^{\text{y}} \bigg[\frac{\text{y}}{\text{x}} + \log \text{x}. \frac{\text{dy}}{\text{dx}}\bigg]$
$\text{x} \log \text{y} = \log \text{v} \Rightarrow \frac{\text{dv}}{\text{dx}} = \text{y}^{\text{x}} \bigg[\frac{\text{x}}{\text{y}} \frac{\text{dy}}{\text{dx}} + \log \text{y}\bigg]$
$\text{putting in (i)} \text{x}^{\text{y}} \bigg[\frac{\text{y}}{\text{x}} + \log \text{x} \frac{\text{dy}}{\text{dx}}\bigg] + \text{y}^{\text{x}} \bigg[\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}} + \log \text{y} \bigg]= 0$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = - \frac{\text{y}^{\text{x}} \log \text{y + y.x}^{\text{y - 1}}}{\text{x}^{\text{y}}. \log \text{x + x.y}^{\text{x - 1}}}$

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