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

 

 

 

 
 
 
 
 

Answers (1)

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}

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