Explaining why $\displaystyle {d \over dt}e^{\int_0^tA(s)\,ds} \ne A(t)e^{\int_0^tA(s)\,ds}$ by definition I've been asked to explain why $\displaystyle {d \over dt}e^{\int_0^tA(s)\,ds} \ne A(t)e^{\int_0^tA(s)\,ds}$ using the definition of $e^A$, where $A$ is an $n\times n$ matrix.
I don't fully understand how they aren't equal. Someone please point out where I went wrong:
$$A(t)e^{\int_0^tA(s)\,ds} = A(t) \sum_{n = 0}^\infty{1 \over n!}\left(\int_0^tA(s)\,ds\right)^n$$
We have been given that 

$${d \over dt}\left(\int_0^tA(s)\,ds\right)^n = nA(t)\left(\int_0^tA(s)\,ds\right)^{n-1},$$ 

which I use to write $$\begin{align}{d \over dt}e^{\int_0^tA(s)\,ds} &= {d\over dt}\left(\sum_{n=0}^\infty{1 \over n!}\left(\int_0^tA(s)\,ds\right)^n\right) \\ & = \sum_{n=0}^\infty{1 \over n!}{d\over dt}\left(\int_0^tA(s)\,ds\right)^n\\&=\sum_{n=1}^\infty{1\over(n-1)!}A(t)\left(\int_0^tA(s)\,ds\right)^{n-1}\end{align}$$
They look equal to me, but likely there's a mistake somewhere. Can anyone see what's wrong?
 A: Notice that, according to Leibniz's rule,
$$\frac {\Bbb d} {\Bbb d t} \left(\int \limits _0 ^t A(s) \ \Bbb d s \right) ^n = \frac {\Bbb d} {\Bbb d t} \underbrace {\int \limits _0 ^t A(s) \ \Bbb d s \dots \int \limits _0 ^t A(s) \ \Bbb d s} _{n \text{ times}} = \color{red} {A(t)} \underbrace{ \int \limits _0 ^t A(s) \ \Bbb d s \dots \int \limits _0 ^t A(s) \ \Bbb d s} _{n-1 \text{ times}} + \int \limits _0 ^t A(s) \ \Bbb d s \color{red} {A(t)} \underbrace{ \int \limits _0 ^t A(s) \ \Bbb d s \dots \int \limits _0 ^t A(s) \ \Bbb d s} _{n-2 \text{ times}} + \dots \underbrace{\int \limits _0 ^t A(s) \ \Bbb d s \dots \int \limits _0 ^t A(s) \ \Bbb d s} _{n-1 \text{ times}} \color{red} {A(t)} ,$$
where $A(t)$ occupies, in turn, all the positions from $1$ to $n$ in that product, and it is not at all obvious why $A(t)$ should commute with $A(s)$ for $0 \le s < t$.
If, on the other hand, these matrices do commute, then indeed, the result is as you have written, namely
$$\color{red} {A(t)} \ n \left( \int \limits _0 ^t A(s) \ \Bbb d s \right) ^{n-1} .$$
A: Calling the whole integral at the exponent as $A$ you get:
$$\frac{\text{d}}{\text{d}t} e^A = \frac{\text{d}}{\text{d}t}\sum_{k = 0}^{+\infty}\frac{A^k}{k!} = \sum_{k = 0}^{+\infty}\frac{1}{k!}\frac{\text{d}}{\text{d}t}(A^k)$$
Now
$$\frac{\text{d}}{\text{d}t} A^k = k\cdot A^{k-1}\cdot \frac{\text{d}A}{\text{d}t}$$
Hence
$$\frac{\text{d}}{\text{d}t} e^A = \sum_{k = 0}^{+\infty}\frac{1}{k!}\frac{\text{d}}{\text{d}t}(A^k) = \sum_{k =0}^{+\infty}\frac{1}{k!}k\cdot A^{k-1}\frac{\text{d}A}{\text{d}t} = A\cdot\frac{\text{d}A}{\text{d}t} \sum_{k = 0}^{+\infty}\frac{k\cdot A^k}{k!}$$
And you can see that it's proven.
