Sorry if I worded the question title weirdly, I'm not sure how to word it.

I used Wolfram Alpha to try to get the derivative of y^x

Here is the input

Here is the output (snapshot), as you can see there is a 3D plot and I get that the x and y axis are literally x and y from the equation, but what is the z-index on this plot? I would imagine it's the imaginary values, yet the plot clearly indicates this is the RE(al) part of the equation and the z-index labels have no "i" indicator in their numbers.

It looks like it's almost a scale for the x and y axis, like the values there are multiplied by the z-axis value. Having a hard time wrapping my head around why one would do that though, if that is the case it's the first time I've seen it but I don't deal with 3D Plots ever haha. Attached below:

enter image description here

  • $\begingroup$ What are you trying to do? - " get the derivative of y^x" - what kind of derivative? Partial derivative with respect to $x$? Or is $y$ a function of $x$? Wolfram Alpha has no idea what are you asking, which is why it tried to take both $x$ and $y$ derivatives $\endgroup$ – Yuriy S Nov 6 '16 at 9:06
  • $\begingroup$ @YuriyS I was initially going to take it w/ respect to change of x but forgot to type that in, before I could correct it I noticed this off 3D plot and had to come ask about it just out of curiosity. The derivative I did end up with was y^x*ln(x) but that's likely irrelevant to this question at this point, I'm just curious about the graph haha $\endgroup$ – Albert Renshaw Nov 6 '16 at 10:18

I think complex numbers are "creeping into" this issue due to the presence of $\log$ function. In case (not shown here) that $y$ takes negative values, you may know that $\log(y)$ is defined, but as a complex entity (for example to $\log(-1)$ can be considered as $log(e^{i \pi})=i \pi$. This is why studying complex function theory is so important).

  • $\begingroup$ This was my first thought but the graph/plot is labeled "Real part" and the whole function is cased in re( ) $\endgroup$ – Albert Renshaw Nov 6 '16 at 10:19

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