Functions that cannot be differentiated in terms of elementary functions A while ago, I learned how to take the derivative of $y=x^x$ using implicit differentiation, and I wondered if the same trick would work on every function of this type. I tried to differentiate $y=x^{x^x}$ the same way:
$\ln y=x^x\ln x$
$\ln (\ln y)=x \ln x+\ln (\ln x)$
$\frac{\frac{dy}{dx}}{y\ln y}=1+\ln x+\frac{1}{x\ln x}$
$\frac{dy}{dx}=x^{x^x+x}\ln x(1+\ln x+\frac{1}{x\ln x})$
This result seems to imply that all functions of this type, no matter how complicated, could be differentiated in this manner. My question is: Are there any functions of this sort that are so complicated that they are impossible to differentiate? If not, can this be proven?
 A: First, your question is about the existence of an algorithm that computes the derivative of a function that can be expressed with addition, product, quotient, power, and trigonometric, exponential and logarithmic functions. This is a question about computability, not about derivability. In anty case, I think that it is a good question.
The operations that I have mentioned: some are unary, some are binary. The unary operations are covered by the chain rule. For the binary ones, we have:
$$(f+g)'=f'+g'$$
$$(fg)'=f'g+fg'$$
$$\left(\frac fg\right)'=\frac{f'g-fg'}{g^2}$$
$$\left(f^g\right)'=f^g\left[g'\ln f+\frac {f'g}f\right]$$
A: Consider any family $\mathfrak{F}$ of all functions $f(x)$ for which $f'(x)$ if expressible in terms of arithmetic operations, exponents, and members of $\mathfrak{F}$.
You can show using all the rules of derivatives that every expression constructed using only arithmetic operations, exponents, and members of $\mathfrak{F}$, must itself belong to $\mathfrak{F}$.
This result guarantees that closed formulas for derivatives of expressions you tried must exist.
On another note, taking derivatives of sufficiently complicated expressions quickly becomes impractical as you can expect the expressions to be monsterous in size.
