TITLE: Find the Jacobian of $f(x,y,z)=x^y$
QUESTION [5 upvotes]: Let $f(x,y,z)=x^y$. I am trying to find the Jacobian $f’(a,b,c)$ without using partial derivatives, i.e. using the Chain Rule and several other elementary results on component functions etc..
Since $x^y=\exp(y\cdot\ln(x))$, using the identity function $\pi$ we can write $$f(x,y,z)=\exp\left(\pi^2 \cdot \ln(\pi^1)\right)(x,y,z).$$
The proof then seems to require the Chain Rule, with $g(x)=\exp(x)$ and $h(x,y,z) =\pi^2 \ln(\pi^1)(x,y,z)$, so that $f=g\circ h $ and $f’(a,b,c)= g’\left(h(a,b,c)\right)\cdot h’(a,b,c)$. From here, I can calculate the solution $$\left[ba^{b-a}\; \; a^b \ln(a)\; \; 0\right].$$
However,in order to apply the Chain Rule, we require $h:\mathbb{R}^n\to \mathbb{R}^p$ and $g: \mathbb{R}^p \to  \mathbb{R}^m $ so that $f: \mathbb{R}^n \to \mathbb{R}^m$ is defined by $f= g\circ h$. To get the right solution, however, I had to use $g’(h(a,b,c))=[a^b]$.

Question: How can we use the function $g:\mathbb{R}\to\mathbb{R}$ in the above proof for $f’(x)$?

REPLY [3 votes]: The function $f:\Bbb R^3\to\Bbb R$ given by $f(x,y,z)=x^y$ can be written as $$f=\exp\circ(\pi^2\cdot\log[\pi ^1]).$$ This gives \begin{align}f'(a,b,c)&=\exp'(b\log a)\cdot[b(\log[\pi^1])'(a,b,c)+(\log a)(\pi^2)'(a,b,c)+0(\pi^3)'(a,b,c)]\\&=\exp(b\log a)\cdot\left[b\cdot a^{-1}(\pi^1)'(a,b,c)+(\log a)(\pi^2)'(a,b,c)+(0,0,0)\right]\\&=a^b\cdot\left[ba^{-1}(1,0,0)+(\log a)(0,1,0)+(0,0,0)\right]\\&=(ba^{b-1},a^b\log a,0).\end{align}
The dimensions are correct for the Chain Rule to be applied. Let $p=1$, $m=1$ and $n=3$. Then we have \begin{align}g&:\Bbb R^p\to\Bbb R^m\to g(x)=\exp x\\h&:\Bbb R^n\to\Bbb R^p\to h(x,y,z)=(\pi^2\cdot\log[\pi^1])(x,y,z)\\f\equiv g\circ h&:\Bbb R^n\to\Bbb R^m\to f(x,y,z)=\exp\circ(\pi^2\cdot\log[\pi^1])(x,y,z).\end{align}