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

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Given: $x^y - y^x = a^b$ , find $\frac{dy}{dx}$

let $x^y= u\qquad y^x = v$

$y\log x = \log u\ \ -(1)\quad x\log y = \log v \ \ -(2)$

differentiating (1) w.r.t to $x$

$\frac{y}{x} + \log x\frac{dy}{dx} = \frac{1}{u}\frac{du}{dx}$

$\frac{du}{dx} = u\left [\frac{y}{x} + \log x\frac{dy}{dx} \right ]$

$=x^y\left [\frac{y}{x} + \log x\frac{dy}{dx} \right ]$

$\frac{du}{dx} = yx^{y-1} + x^y\log x\frac{dy}{dx} \ \ -(3)$

DIfferentiating (2) w.r.t to $x$

$\frac{x}{y}\frac{dy}{dx} + \log y = \frac{1}{v}\frac{dv}{dx}$

$\frac{dv}{dx} = v\left [\frac{x}{y}\frac{dy}{dx} + \log y \right ]$

$\frac{dv}{dx} = xy^{x-1}\frac{dy}{dx} + y^x\log y \ \ -(4)$

$\Rightarrow x^y - y^x = a^b$

$u -v =a^b$

Differentiating w.r.t $x$

$\frac{du}{dx} -\frac{dv}{dx} = 0$

From (3) and (4)

$yx^{y-1} + x^y\log x\frac{dy}{dx} - \left[xy^{x-1}\frac{dy}{dx} + y^x\log y \right ]=0$

$\frac{dy}{dx} \left[x^y\log x +y^{x-1}x \right ] = \left[y^x\log y + yx^{y-1} \right ]$

$\frac{dy}{dx} = \frac{y^x\log y + yx^{y-1}}{x^y\log x +y^{x-1}x}$

Posted by

Ravindra Pindel

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