Question

Find derivative of
${x}^{y} * {y}^{x} = 12$
​

Answer 1

$\large\underline{\sf{Formula \: Used-}}$

$\bf \: \dfrac{d}{dx} x = 1$

$\bf \: \dfrac{d}{dx} {e}^{x} = {e}^{x}$

$\bf \: \dfrac{d}{dx} logx = \dfrac{1}{x}$

$\bf \: \dfrac{d}{dx} u.v = v \: \dfrac{d}{dx} u \: + \: u \: \dfrac{d}{dx} v$

$\bf \: {x}^{y} = {e}^{y \: logx}$

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\: {x}^{y} + {y}^{x} = 12$

$\rm :\longmapsto\: {e}^{ylogx} + {e}^{xlogy} = 12$

On differentiating w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}( {e}^{ylogx} + {e}^{xlogy}) =\dfrac{d}{dx} 12$

$\rm :\longmapsto\: {e}^{ylogx} \dfrac{d}{dx} (ylogx) + {e}^{xlogy} \dfrac{d}{dx} (xlogy) = 0$

$\rm :\longmapsto\: {e}^{ylogx} (y\dfrac{d}{dx} logx + logx\dfrac{d}{dx} y) + {e}^{xlogy} (x\dfrac{d}{dx} logy + logy\dfrac{d}{dx} x) = 0$

$\rm :\longmapsto\: \sf \: {x}^{y} \bigg(\dfrac{y}{x} + logx\dfrac{dy}{dx} \bigg) + {y}^{x}\bigg(\dfrac{x}{y}\dfrac{dy}{dx} + logy \bigg) = 0$

$\rm :\longmapsto\: y{x}^{y - 1} + {x}^{y} logy\dfrac{dy}{dx} + {xy}^{x - 1} \dfrac{dy}{dx} + {y}^{x} logy = 0$

$\rm :\longmapsto\: {x}^{y} logy\dfrac{dy}{dx} + {xy}^{x - 1} \dfrac{dy}{dx} = - {y}^{x} logy - y{x}^{y - 1}$

$\rm :\longmapsto\: ({x}^{y} logy + {xy}^{x - 1}) \dfrac{dy}{dx} = - ( {y}^{x} logy + y{x}^{y - 1} )$

$\rm :\longmapsto\: \dfrac{dy}{dx} = \dfrac{ - ( {y}^{x} logy + y{x}^{y - 1} )}{ ({x}^{y} logy + {xy}^{x - 1})}$

$\rm :\longmapsto\: \dfrac{dy}{dx} = \dfrac{ - y( {y}^{x - 1} logy + {x}^{y - 1} )}{x ({x}^{y - 1} logy + {y}^{x - 1})}$