why exactly doesn't logarithmic differentiation work here? As part of a larger problem, I have to differentiate $y^x$
so if we let $y=y^x$
we get $\ln(y)=x\ln(y)$
which means the differential is zero
alternatively, let $y=e^{\ln(x^y)}$
so $y=e^{x\ln(y)}$
which can be differentiated as follows
$y'=x^y\left(\ln(y)+y'\ \frac xy\right)$
which is $y'\left(1-{(x^y)}\ \frac xy\right)=x^{y}\ln(y)$
which means $y'=x^y\ln\left(\frac {y}{\left(1- {(x^y)}\frac{x}{y}\right)}\right)$
I apologise for not using MathJaX. I can't seem to get it to work for me
 A: Differentiating the function $f(x) = y^x$ (for some $y$ which is a function of $x$) is not the same thing as using implicit differentiation on $y=y^x$.
For example, if $y=e^x$, then $f(x)=y^x$ is the function $f(x)=(e^x)^x = e^{x^2}$; but $y=y^x$ gives $e^x=e^{x^2}$, which gives $e^{x^2-x}=1$, so the only values of $x$ that satisfy the equation are $x=0$ and $x=1$ and the graph of $y=y^x$ consists of exactly two points, $(0,1)$ and $(1,e)$.
The graph of $y=y^x$ consists of all points of the form $(x,0)$ with $x\gt 0$; the line $y=1$; and the line $x=1$. Indeed, all of these satisfy the equation: if $y=0$, then we need $x\gt 0$ for $y^x$ to make sense, and we get $0=0^x$; if $y=1$ then we get $1=1^x$, which holds for all $x$; and if $x=1$ we get $y=y^1$ which holds for all $y$. And these are the only solutions. If $y=0$, then we must have $x\gt 0$ (the positive $x$-axis); if $y\neq 0$, then $y^{x-1}=1$, so either $y=1$ and $x$ is arbitrary (the line $y=1$); or else $x-1=0$ (the line $x=1$ with $y\neq 0$).
Here, the derivative exists only on the positive $x$ axis and the line $y=1$. Of those, we can only use logarithmic differentiation for $y=1$, where we get
$\ln |y| = \ln|y^x|$, or $\frac{y'}{y} = \frac{xy'}{y}$, or $y'(x-1)=0$. This means $y'=0$, as expected.
The main problem with what you are doing is that you are trying to differentiate $f(x) = y^x$, but you are actually finding the implicit derivative for $y=y^x$, and there only for values with $y\neq 0$... where you just get $y'=0$ because of what that graph is.
You can use logarithmic differentiation for $y^x$ (leaving $y'$ indicated) if you set it up correctly. Setting $f(x)=y^x$, we have:
$$\begin{align*}
\ln |f(x)| &= \ln |y^x|\\
\ln |f(x)| &= x\ln|y|\\
\frac{d}{dx} \ln|f(x)| &= \frac{d}{dx}\left(x\ln|y|\right)\\
\frac{f'(x)}{f(x)} &= \ln|y| + x\left(\frac{y'}{y}\right)\\
f'(x) &= f(x)\left(\ln|y| + x\frac{y'}{y}\right)\\
f'(x) &= y^x\left(\ln|y| + x\frac{y'}{y}\right)\\
f'(x) &= (\ln|y|)y^x + xy^{x-1}y'\\
f'(x) &= (\ln y)y^x + xy^{x-1}y',
\end{align*}$$
with last equality because $y^x$ is only defined for $y\gt 0$.
If you do it directly using $a^b=e^{b\ln a}$, we have:
$$\begin{align*}
\frac{d}{dx}y^x &= \frac{d}{dx} e^{x\ln y}\\
&= (x\ln y)'e^{x\ln y}\\
&= \left(\ln y + x\frac{y'}{y}\right) e^{x\ln y}\\
&= \left( \ln y + x\frac{y'}{y}\right)y^x\\
&= (\ln y)y^x + xy^{x-1}y',
\end{align*}$$
exactly the same result as before.
