6
$\begingroup$

Need to differentiate following equation: $$x^y=y^x$$ My attempt: $$x^y\log(x) \cdot y' = y^x\log(y)\cdot1; $$ $$y'=\frac{y^x\log(y)}{x^y\log(x)}$$ Please tell me if I've made a mistake.

$\endgroup$
  • 2
    $\begingroup$ I think he means "take the derivative." $\endgroup$ – Thomas Andrews Nov 1 '13 at 20:51
  • 3
    $\begingroup$ "$x^y=y^x$ is not an expression, but an equation defining a solution set in the first quadrant of the $(x,y)$-plane. What do you mean by "deriving" this equation? $\endgroup$ – Christian Blatter Nov 1 '13 at 21:00
  • 1
    $\begingroup$ I have suggested the edit. $\endgroup$ – Ahaan S. Rungta Nov 1 '13 at 21:05
  • $\begingroup$ @ChristianBlatter I really don't know what I mean $\endgroup$ – k1ber Nov 1 '13 at 21:08
11
$\begingroup$

Not quite. The idea is to take the $\log$ of both sides and implicitly differentiate. $$ \begin {eqnarray*} x^y &=& y^x \\ y \cdot \log (x) &=& x \cdot \log (y) \\ y' \cdot \log(x) + \dfrac {y}{x} &=& \dfrac {x}{y} \cdot y' + \log (y) \\ y' \cdot \left( \log(x) - \dfrac {x}{y} \right) &=& \log(y) - \dfrac {y}{x} \\ y' &=& \dfrac {\log (y) - \dfrac {y}{x}}{\log (x) - \dfrac {x}{y}}, \end {eqnarray*} $$so our answer is $$ y' = \boxed {\dfrac {x \cdot \log (y) - y}{y \cdot \log (x) - x} \cdot \dfrac {y}{x}}. $$

$\endgroup$
  • 3
    $\begingroup$ Answer should be multiplied by y/x, x being the denominator of the numerator and y being the denominator of the denominator in the step leadinf to "so our answer is". $\endgroup$ – Bernard Massé Nov 21 '14 at 21:23
  • $\begingroup$ @BernardMassé - whoops, you're right - thanks! Fixed. :) $\endgroup$ – Ahaan S. Rungta Nov 21 '14 at 23:41
9
$\begingroup$

Just to let you know where you went wrong: it was in differentiating the right hand side. You're trying to find

$ \frac{d}{dx} y^x $

and applying the rule that

$ \frac{d}{dx} a^x = a^x \ln (a), $

with $y$ replacing $a$. Unfortunately, that "rule" applies only in the case where $a$ is a constant; in your problem, the base $y$ is actually a function of $x$, and so the rule doesn't apply. Thus to properly compute the derivative, you have to rewrite $y^x = e^{x \ln (y)}$, as others have already shown.

$\endgroup$
-1
$\begingroup$

$ y^x = x^y $

$ \ln y^x = \ln x^y$

$x \ln y = y \ln x $

$\ln y + (x/y)(dy/dx) = (dy/dx)(ln x) + y/x $

$(x/y)(dy/dx) - (dy/dx)(\ln x) = y/x - \ln y $

$(dy/dx)(x/y - \ln x) = y/x - \ln y $

$(dy/dx) = (y/x - \ln y)/(x/y - \ln x)$

$(dy/dx) = (y^2 - xy\ln y)/(x^2 - xy\ln x)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.