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

Question

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

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Answer

${\text{x}}^{\text{x}} + \text{x}^{\text{y}} + \text{y}^{\text{x}} = \text{a}^{\text{b}}$
$\text{Let u = x}^{\text{x}},\text{v = x}^{\text{y}}, \text{w = y}^{\text{x}}, \frac{\text{du}}{\text{dx}} + \frac{\text{dv}}{\text{dx}} + \frac{\text{dw}}{\text{dx}} = 0$
$\frac{\text{du}}{\text{dx}} = \text{x}^{\text{x}} (1 + \log \text{x})$
$\frac{\text{dv}}{\text{dx}} = \text{x}^{\text{y}} \bigg(\frac{\text{y}}{\text{x}} + \frac{\text{dy}}{\text{dx}}\log \text{x}\bigg)$
$\frac{\text{dw}}{\text{dx}} = \text{y}^{\text{x}}\bigg(\frac{\text{x}}{\text{y}}. \frac{\text{dy}}{\text{dx}} + \log \text{y}\bigg)$
$\frac{\text{dy}}{\text{dx}} = \bigg(\frac{\text{x}^{\text{x}}( 1 + \log \text{x}) + \text{y x}^{\text{y} - 1} + \text{y}^{\text{x}} \log \text{y}}{\text{x}^{\text{y}} \log \text{x} + \text{x y}^{\text{x} - 1}}\bigg)$

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