Can someone tell me the correct partial derivative of $x^{y^z}$ on the variable z. I got two solutions:
$x^{y^z}\ln\left(x^y\right)$
and
$x^{y^z}y^z\ln x\ln y$
They both seem right to me so I am confused.
Which one is correct and why?
[Math] partial derivative of $x^{y^z}$
derivatives
Related Solutions
$\def\part#1#2{{\partial#1\over\partial#2}}$ $\def\parts#1#2#3{{\partial^2#1\over\partial#2\,\partial#3}}$
On the left hand side of your equations, you have the symbol "$\part{\vphantom f}y\bigl(\part f x\bigr)"$. By definition this is the partial derivative of the function $\part fx$ with respect to $y$. So, upon encountering this symbol, you take the function $\part fx$ and then take its partial with respect to $y$. The natural notation of the type on the right hand side of your equations is the notation used in (2) of your post: $$\tag{3} \part{\vphantom f}y\Bigl(\part f x\Bigr)=\parts f y x. $$
I will not surmise why this is the "natural" notation, but will point out that $(3)$ gives the adopted definition for the symbol $\parts f y x$ in any calculus/analysis text, or any other "credible" source, you'll find.
I emphasise here that $(3)$ defines the symbol $\parts f y x$; that this sometimes gives an expression that equals $\parts f x y$ is irrelevant. (Of course, for certain functions, what you wrote in (1) would be correct; but its correctness would follow from the result of a theorem, not from the definition of the symbols.)
I dislike using the same notation for both a random variable and the argument to its density function or cumulative distribution function. Thus $F_X(x)=\Pr(X\le x)$, and the meanings of $X$ and $x$ are different. The lower-case $x$ can be any number, so for example $F_X(3)=\Pr(X\le3)$, etc. I think a good case can be made that certain notational usages ultimately make clear thinking easier. In particular, $f_Y(y)$ is not a random variable, but $f_Y(Y)$ is, and it looks as if that's what we've got here. So let's say $$ Y\mid\theta \sim N(\theta,1)\text{ and }\theta\sim N(0,\delta^2). $$
I take it $P(y\mid\theta)$ is supposed to be the conditional density of $Y$ given $\theta$ evaluated at $y$. I'll write $f_{Y\mid\theta}(y)$.
Everything you're asking about is conditional on the value of $\theta$ (or did I miss something?) so the distribution of $\theta$ doesn't matter at all; for the purposes of this problem it's not even a random variable. I don't know what a partial derivative with respect to a random variable would mean (except maybe a Radon--Nikodym derivative, but that has no relevance here).
Now we have $$ \mathbb E\left[ -\frac{\partial^2}{\partial\theta^2} \ln f_{Y\mid \theta}(Y) \mid\theta \right]. $$ $$ = \mathbb E \left[ -\frac{\partial^2}{\partial\theta^2} \ln\left( \frac{1}{\sqrt{2\pi}} \exp\left( \frac{-(Y-\theta)^2}{2} \right) \right) \right] =\mathbb E\left[ -\frac{\partial^2}{\partial\theta^2}\left( \frac{-(Y-\theta)^2}{2} \right) \right]. $$ Then it appears you have the expected value of a constant.
Best Answer
If $$f(x,y,z) = x^{y^{z}} \text{ then } \frac{\partial}{\partial z} f(x,y,z) = \log(x) y^z \log y x^{y^{z}}$$ And if, $$f(x,y,z) = (x^y)^{z} \text{ then } \frac{\partial}{\partial z} f(x,y,z) = (x^y)^{z} \log(x^y)$$ These results are different due to the different ways the power of $z$ has been executed. Hope it helps.