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$$x^y = y^x$$

find an expression for $dy/dx$ in terms of $y$ and $x$

for the first part is $yx^{y-1}$ and for the second part i got $\ln(y)y^xdy/dx$

when solving for dy/dx i got $yx^{y-1}) / (\ln(y)y^x)$

Just was wondering if my work is correct?

Sorry for the bad formatting, Im not sure how to make it formatted nice.

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what do you do to get the formatting of the exponents? –  user71317 Apr 6 '13 at 23:02
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1 Answer

up vote 3 down vote accepted

It might be easier to take $\log$ first and work with the expression. We have $$y \log x = x \log y$$ Now differentiating we get $$y' \log x +\dfrac{y}x = \log y + x \dfrac{y'}y \implies y' \left(\log x - \dfrac{x}y \right) = \left(\log y - \dfrac{y}x \right)$$ Hence, we get that $$y' = \dfrac{\left(\log y - \dfrac{y}x \right)}{\left(\log x - \dfrac{x}y \right)}$$

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