Derivative of Complex Exponential I would like help with this problem.
Prove this is true by using complex analysis. This problem appeared in my workbook and there appears to be no solution in the back. 
$$\frac{d}{dx}e^{tx} = te^{tx}.$$
Edit: I think I am supposed to use chain rule? 
 A: If you don't mind, I will use the complex number letters I am used to.
So a complex function of the complex variable is holomorphic at $z_0$ if 
$$
\lim_{z\rightarrow z_0}\frac{f(z)-f(z_0)}{z-z_0}
$$
exists in $\mathbb{C}$. In this case, it is denoted by $f'(z_0)$. 
This notion obeys the usual rules you are used to for derivatives of real functions of the real variable. In particular, there is a chain rule. But note there is much much more structure on holomorphic functions, than on differentiable real functions of the real variable. 
So the function you are considering is
$$
g(z):=e^{uz}.
$$
This is the composition $g=\exp\circ f$ with $f(z)=uz$. Now $\exp$ is holomorphic on $\mathbb{C}$ with derivative equal to itself (this is a fact which follows from the definition of $\exp$ as the sum of the series $\sum_{n\geq 0} z^n/n!$), and $f$ is holomorphic on $\mathbb{C}$ with derivative equal to $u$ (this is easy from the definition).
Now by the chain rule, $g$ is holomorphic on $\mathbb{C}$ and
$$
g'(z)=\exp'(f(z))f'(z)=e^{uz}u.
$$
