What is the derivative of y when x^y = y^x?

Can someone please give me the steps to solving this equation?

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- Oct 22nd 2009, 08:54 PMamma0913x^y = y^x
What is the derivative of y when x^y = y^x?

Can someone please give me the steps to solving this equation? - Oct 22nd 2009, 10:12 PMcalum
The first step is to take the natural log of both sides:

$\displaystyle ln(x^y) = ln(y^x) \therefore yln(x) = xln(y)$

Taking the derivative of both sides with respect to x, using the product and chain rules, gets:

$\displaystyle y \frac{d}{dx}ln(x) + ln(x){dy}{dx}= x \frac {d}{dx}ln(y) + ln(y){d}{dx}x$

$\displaystyle \frac{y}{x} + ln(x) \frac{dy}{dx}= x \frac {d}{dy}ln(y) \frac{dy}{dx} + ln(y)$

$\displaystyle \frac{y}{x} + ln(x) \frac{dy}{dx} = \frac{x}{y} \frac {dy}{dx} + ln(y)$

$\displaystyle \frac{dy}{dx} = \frac{ln(y) - \frac{y}{x}}{ln(x) - \frac{x}{y}}$