# Differentiating an exponential function

I'm trying to calculate a Christoffel symbol, but I'm stuck on showing that $$\frac{\partial(e^{2A})}{\partial r}=2e^{2A}\frac{\partial A}{\partial r}\ ?$$

Nice easy steps would be much appreciated.

Thank you

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What exactly is the problem? This is the chain rule. Assumming $A = A(r)$, $A$ is differentiable w.r. to $r$ and $e^x$ is the common exponential function. – user20266 Dec 5 '11 at 18:40
@Peter4075: Good! Then you've got it! Just take $u=2A$ in your case. You will have $du/dr=2 dA/dr$, and that's all there is to it. – Hans Lundmark Dec 5 '11 at 19:23
Dumb question: Is this a matrix exponential? Is that why you're worried? Best, – Dylan Moreland Dec 5 '11 at 20:32
Another dumb question then: how come you're calculating Christoffel symbols...? No offence, but that sounds quite advanced if you're having trouble with the chain rule. (P.S. For messages to another user, put "@" before the user name, and they will get notification in their inbox at the upper left of the page.) – Hans Lundmark Dec 5 '11 at 21:41
And by the way: perhaps if you write $u=A(r)$ it's easier to see why $r$ is the variable to differentiate with respect to; $A(r)$ just symbolizes any function of $r$, like $A(r)=r^5$ in the example. – Hans Lundmark Dec 5 '11 at 21:43

When $A$ is a scalar function of $r$, the formula $\frac{d }{d r}e^{2A}=2e^{2A}\frac{dA}{dr}$ is nothing but the chain rule.
When $A$ is a matrix, one has to watch out for the lack of commutativity. If $A$ and $\frac{dA}{dr}$ commute, everything works as for scalars. Otherwise, the formula breaks down: for example, if $$A(r)=\begin{pmatrix}0 & 1 \\ r & 0\end{pmatrix} \tag1$$ then $$\frac{d}{dr}_{\, \big|r=0} \exp A(r) = \begin{pmatrix}1/2 & 1/6 \\ 1 & 1/2\end{pmatrix}\tag2$$ which is not a product of $A'(0)=\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}$ and $\exp A(0)=\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$, in either order: $$A'(0)\exp A(0) =\begin{pmatrix}0 & 0 \\ 1 & 1\end{pmatrix}\qquad (\exp A(0)) \,A'(0)=\begin{pmatrix}1 & 0 \\ 1 & 0\end{pmatrix}\tag3$$ One may observe in passing that the average of two products in (3) is $$\frac12 (A'(0)\exp A(0) + (\exp A(0)) \,A'(0)) = \begin{pmatrix}1/2 & 0 \\ 1 & 1/2\end{pmatrix} \tag4$$ which does a decent job of approximating (2). This suggests that (4) may be the beginning of a power series that gives the correct value of the derivative of matrix exponential, which (I guess) can be fished out of the Zassenhaus formula.