Alternative methods to obtain the derivative of an elementary function (besides the definition, and the sum/product/chain rules) Given a differentiable function $f:\mathbb R \to \mathbb R$ which is an elementary function, what methods are there to calculate the its derivative?
(1) We of cause have the possibility of using the definition itself:
$$
f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
$$
(2) We define some basic functions for which we look up their derivative, and use sum, product, and chain rule to calculate $f'(x)$.
What other methods are there? 
 A: We could numerically approximate the derivative. Most numerical methods rely on manipulating Taylor series expansions.
(1) The most common of these are finite difference methods. Taylor expansions allow us to write the forward difference, backward difference, and central difference. 
Given a small $h > 0$, we have 
Forward Difference: Starting from 
$$f(x+h) = f(x)+hf'(x)+\frac{1}{2!}h^2f''(x)+\frac{1}{3!}h^3f'''(x)+\frac{1}{4!}h^4f''''(x)+\dots$$
move $f(x)$ to the LHS and divide by $h$ to form
$$f'(x)\approx \frac{f(x+h)-f(x)}{h}$$
Backward Difference: Starting from
$$f(x-h) = f(x)-hf'(x)+\frac{1}{2!}h^2f''(x)-\frac{1}{3!}h^3f'''(x)+\frac{1}{4!}h^4f''''(x)-\dots$$
move $f(x-h)$ to the RHS, $f'(x)$ to the LHS, and divide by $h$ to form
$$f'(x)\approx \frac{f(x)-f(x-h)}{h}$$
Central Difference: Starting from
$$f(x+h) = f(x)+hf'(x)+\frac{1}{2!}h^2f''(x)+\frac{1}{3!}h^3f'''(x)+\frac{1}{4!}h^4f''''(x)+\dots$$
subtract
$$f(x-h) = f(x)-hf'(x)+\frac{1}{2!}h^2f''(x)-\frac{1}{3!}h^3f'''(x)+\frac{1}{4!}h^4f''''(x)-\dots$$
and then divide by $2h$ to form
$$f'(x)\approx \frac{f(x+h)-f(x-h)}{2h}$$
Higher order derivatives are known and this book is a great introduction to the theory of the finite difference method.
(2) We can apply the method of undetermined coefficients to generate approximations to derivatives. 
Suppose $$f'(x)\approx D_hf(x)\equiv Af(x-h) + Bf(x)+Cf(x+h)\tag{1}$$
where $A,B,C$ are constants that we must determine so that $D_hf(x)$ is as accurate as possible. Then, from the Taylor series,
$$f(x-h) = f(x)-hf'(x)+\frac{1}{2!}h^2f''(x)-\frac{1}{3!}h^3f'''(x)+\frac{1}{4!}h^4f''''(x)-\dots\tag{2}$$
$$f(x)=f(x)\tag{3}$$
$$f(x+h) = f(x)+hf'(x)+\frac{1}{2!}h^2f''(x)+\frac{1}{3!}h^3f'''(x)+\frac{1}{4!}h^4f''''(x)+\dots\tag{4}$$
If we multiply $(2),(3),(4)$ by $A,B,C$ and then add, we see from $(1)$ that
\begin{align}
D_hf(x)&=(A+B+C)f(x)+(C-A)hf'(x) + (A+C)\frac{1}{2!}h^2f''(x) + (C-A)\frac{1}{3!}h^3f'''(x)\\&~~~~~+(A+C)\frac{1}{4!}h^4f''''(x)+\dots
\end{align}
Therefore, if
$$D_hf(x)\approx f'(x)$$
for some arbitrary function $f(x)$, we would require that
$$A+B+C=0$$
$$h(C-A)=1$$
$$A+C=0$$
If we solve this system, we will obtain
$$A=-C=-\frac{1}{2h}$$
$$B=0$$
which would then produce our central difference scheme
$$D_hf(x)\equiv \frac{f(x+h)-f(x-h)}{2h}$$
and
$$D_hf(x)=f'(x)+\frac{1}{3!}h^2f'''(x)+\dots$$
We can also change the coefficients of $A,B,C$ so that $D_hf(x)$ is exact for polynomials of as high degree as possible. This will lead to the well known central difference formula to the second derivative
$$D_h^{(2)}f(x)=\frac{f(x-h)-2f(x)+f(x+h)}{h^2}$$
A: By definition, an elementary function is some finite expression containing certain "basic elementary functions" and operations $+$, $\ldots$, exponentiation, inverse functions, etc. The derivatives of the basic  elementary functions are found from their defining properties and your $(1)$, and the derivative of a "finite expression" containing such such basic functions is found using the well known derivative rules on the "expressional tree" defining such a function. Apart from the derivatives of the basic functions all this is in terms of finite expressions, and requires no further computations of limits, let alone power series or integrals.
The answer to your question therefore is: There are none. But of course there are identities among the basic functions, like $\cos^2 x +\sin^2 x\equiv1$, and such identities may allow to simplify the obtained derivatives. This is a question of "normalization", like replacing $2+3$ by $5$, but is not a "new method" of computing derivatives.
