Question

Given that x^y + y^x = a^b, where a and b are positive constants, find ${\displaystyle \frac{dy}{dx}}$.

Answer 1

Given, x^y + y^x = a^b

${\displaystyle e^{{\log x}^y}+e^{{\log y}^x}=a^b}$   ...[we know that: ${\displaystyle e^{\log m}=m]}$

${\displaystyle e^{y\log x}+e^{x\log y}=a^b}$   ...[log mⁿ = n log m]

d.w.r to x

${\displaystyle e^{y\log x}[y\times \frac{1}{x}+\log x\frac{dy}{dx}]+e^{x\log y}[x\times \frac{1}{y}\frac{dy}{dx}+\log y\times 1]=\frac{d}{dx}{a}^b}$

${\displaystyle e{\log x}^y[\frac{y}{x}+\log x\frac{dy}{dx}]+e^{{\log y}^x}[\frac{x}{y}\frac{dy}{dx}+\log y]=0}$

${\displaystyle x^y[\frac{y}{x}+\log x\frac{dy}{dx}]+y^x[\frac{x}{y}\frac{dy}{dx}+\log y]=0}$

${\displaystyle y{x}^{y-1}+x^y\log x\frac{dy}{dx}+x{y}^{x-1}\frac{dy}{dx}+y^x\log y=0}$

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

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