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I know that they probably treated $\displaystyle f(s,t) = e^{-st} f(t)$ so the integration/differentiation thing doesn't matter, but what confuses me is when they got rid of the derivative and how the "$t$" pop out? It's taking the derivative with respect to $s$ not $t$.

Proof: Consider the identity $$\frac{dF(s)}{ds} = \frac{d}{ds} \int_0^{\infty} e^{-st} f(t) dt.$$ Because of the assumptions on $f(t)$, we can apply a theorem from advanced calculus (sometimes called Leibniz's rule) to interchange the order of integration and differentiation: $$ \begin{align} \frac{dF(s)}{ds} & = \frac{d}{ds} \int_0^{\infty} e^{-st} f(t) dt\\ & = \int_0^{\infty} \frac{d \left(e^{-st} \right)}{ds} f(t) dt\\ & = - \frac{d}{ds} \int_0^{\infty} t e^{-st} f(t) dt\\ & = - \mathcal{L} \{ tf(t) \}(s). \end{align} $$ Thus, $$\mathcal{L} \{ tf(t) \}(s) = (-1) \frac{dF(s)}{ds}$$

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And the $t$ then is considered a constant. The chain rule... e.g. $(e^{2s})'=e^{2s}\cdot2$. –  David Mitra May 8 '12 at 3:42
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What happens if you differentiate $e^{-s t}$ with respect to $s$? –  J. M. May 8 '12 at 3:42
    
Thank you for catching my minor brain bug everyone...now my rep is going down lol –  Hawk May 8 '12 at 3:44
    
P.S. you'll sometimes see "differentiation under the integral sign" instead of "Leibniz's rule" in some contexts. –  J. M. May 8 '12 at 3:46

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up vote 2 down vote accepted

In general, $\frac{d}{dx} e^{ax} = a e^{ax}$

Thus, since $t$ is independent of $s$, $\frac{d}{ds} e^{-st} = -te^{-st}$

$f(t)$ doesn't depend on $s$, so it's a constant with respect to differentiation by $s$, and they pulled the negative outside the integral.

Does that make sense?

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Put another way: differentiating with respect to $s$ did not change the fact that $t$ was still a dummy variable within the Laplace transform integral... –  J. M. May 8 '12 at 3:44

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