Whyi can't figure out how they came to this simplification.

[math]x^y = e^{ylnx}[/math]

x^y = exp (y*ln(x) )

>>8688262
How? I'm missing something completely obvious.

>>8688275
[math]\log (x^y) = y \log x [/math]

>>8688275
x=e^lnx so x^y=(e^lnx)^y=e^(lnx*y). When you exponentiate something that already has a power you multiply the two powers

>>8688279
But why are we taking things to an exponential?

Fuck I'm really missing the point here.

>>8688281
Oh okay. Duh. Makes complete sense.

Last question, how does that come up in the differential, then?

>>8688282
So you can directly apply the chain rule.

Think of a simpler example. How do you differentiate [math]f(x) = x^x[/math]? You re-write [math]x^x = \exp(\log(x^x)) = \exp(x \log x)[/math]

>>8688286
Oh sorry I didn't properly read the question. Since z=f(x,y) and both x and y are functions of time to find dz we use the multivariable chain rule

dz=dz/dx*dx+dz/dy*dy

Note these are partial derivatives. i.e. dz/dx means take the derivative of z=x^y while assuming y is constant which gives yx^y-1 and dz/dy=x^y*lnx.

File: image.jpg (2MB, 4032x3024px) Image search: [Google]

2MB, 4032x3024px

>>8688313
I totally forgot about what dz actually means. Going from the formula I worked it out and it makes sense.

Understanding what exactly dz and deltaz are are completely different things though. Still unsure myself.