I know that $e^{i\pi}$ =-1
the derivative of $e^{i\pi}$ is the derivative of - 1 which is 0.
I guess I'm missing a rule or understood something wrong.
$\endgroup$ 14 Answers
$\begingroup$The concept of derivatives is about functions. So, since the derivative of $e^x$ is $e^x$, the derivative of $e^x$ at $i\pi$ is $e^{i\pi}=-1$.
But if you mean the constant function $x\mapsto e^{\pi i}$, its derivative is $0$ everywhere.
$\endgroup$ 1 $\begingroup$The derivative of $e^{ix}$ with respect to $x$ is $ie^{ix}$. You can check this directly by differentiating $\cos(x)+i\sin(x)$.
It turns out that the derivative $e^z$ is $e^z$ also for complex values of $z$, but this may require some knowledge of complex analysis.
$\endgroup$ $\begingroup$$e^{i\pi} = -1$
$e^{i\pi}$ is constant number while $e^{ix} $ is a function. The derivative of $e^{ix}$ at $ x = \pi$ is $ie^{i\pi} = -i.$
But clearly, $e^{i\pi}$ is a constant, so it's derivative is $0.$
This result clearly follows from the definition of derivative. $dy\over dx$ is the rate of change of $y$ with respect to $x$. As the name suggest a constant doesn't depend upon any parameters and hence doesn't change at all. So, the derivative of a constant is $0.$
$\endgroup$ $\begingroup$$$(e^{ax})'=ae^{ax}$$ for any constant $a$, even complex (the prime $'$ denotes differentiation on $x$).
But
$$(e^a)'=0.$$
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