Differentiation with Laplace Transform

Differentiation is related to laplace transform in two ways.

Time domain differentiation (taking laplace of a derivative $f'(t)$ )
Frequency domain differentiation (differentiating the laplaced one itself using $s$ )

Transform of a Derivative (Time Domain)

For a continuous function $f(t)$ that is of exponential order, the Laplace transform of its first derivative $f'(t)$ is given by:

\mathcal{L}\{f'(t)\} = s F(s) - f(0)

where $F(s) = \mathcal{L}\{f(t)\}$ and $f(0)$ is the initial value of $f(t)$ at $t = 0$ .

Exponential Order - Laplace Transform#Existence

Proof via Integration by Parts

Using the definition of the unilateral Laplace transform:

\mathcal{L}\{f'(t)\} = \int_{0}^{\infty} e^{-st} f'(t) \, dt

Apply integration by parts with $u = e^{-st}$ ( $du = -s e^{-st} dt$ ) and $dv = f'(t) dt$ ( $v = f(t)$ ):

\mathcal{L}\{f'(t)\} = \left[ e^{-st} f(t) \right]_{0}^{\infty} - \int_{0}^{\infty} (-s e^{-st}) f(t) \, dt

Assuming $s > 0$ is large enough so that $\lim_{t \to \infty} e^{-st} f(t) = 0$ :

\mathcal{L}\{f'(t)\} = \Big( 0 - f(0) \Big) + s \int_{0}^{\infty} e^{-st} f(t) \, dt = s F(s) - f(0)

Higher-Order Derivatives

Repeatedly applying the first-derivative rule yields the formulas for higher-order derivatives:

Second Derivative:

\mathcal{L}\{f''(t)\} = s^2 F(s) - s f(0) - f'(0)

Third Derivative:

\mathcal{L}\{f'''(t)\} = s^3 F(s) - s^2 f(0) - s f'(0) - f''(0)

Explanation: 
$$
\begin{align*}
\mathcal{L}\{f''(t)\} &= s\left[\mathcal{L}\{f'(t)\}\right] - f'(0) \\ 
					  &= s\left[s\cdot \mathcal{L}\{f(t)\} - f(0) \right] - f'(0) \\ 
					  &= s^2 F(s) - s f(0) - f'(0)
					  
\end{align*}
$$

Recursive Relation:

\text{laplace of N derivative} = s\cdot(\text{laplace of N-1 derivative}) - (\text{N-1 derivative Initial value})

Differentiation in the Frequency Domain ( $s$ -Domain)

Multiplying a function by $t$ in the time domain corresponds to differentiating its Laplace transform in the frequency domain:

\mathcal{L}\{t \cdot f(t)\} = -\frac{d}{ds} F(s)

For any integer power $n \ge 1$ :

\mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s)

$(-1)^n$ happens because of the $e^{-st}$ is differentiated over and over.

Proof

Differentiating $F(s) = \int_0^\infty e^{-st} f(t) \, dt$ with respect to $s$ under the integral sign gives:

\begin{align*} \frac{d}{ds} F(s) &= \frac{d}{ds} \int_{0}^{\infty} e^{-st} f(t) \, dt \\ &= \int_{0}^{\infty} \frac{\partial}{\partial s} \left( e^{-st} \right) f(t) \, dt \\ &= \int_{0}^{\infty} (-t) e^{-st} f(t) \, dt \\ &= -\mathcal{L}\{t f(t)\} \end{align*}