Laplace of a Derivative

The reason the Laplace transform is worth learning is that it converts differentiation into multiplication by $s$:

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

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

and in general

$\mathcal{L}\{f^{(n)}(t)\} = s^nF(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0)$

The hypotheses matter: $f$ must be continuous on $[0,\infty)$, $f'$ piecewise continuous, and both of exponential order. If $f$ jumps at $t = 0$, the rule needs $f(0^+)$ and an extra term.

The boundary values are not optional decoration — they arise from integrating by parts, and they are exactly how initial conditions enter a differential equation.

A related but different property runs the other way:

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

Transforming an equation

1. Write $Y(s) = \mathcal{L}\{y(t)\}$.
2. Replace $y'$ with $sY - y(0)$ and $y''$ with $s^2Y - sy(0) - y'(0)$.
3. Transform the right-hand side from the table.
4. Solve the resulting algebraic equation for $Y(s)$.
5. Invert to get $y(t)$.

The whole point is step 4: a differential equation becomes a linear equation.

Useful transforms for the right-hand side

Checking a rule

Any derivative rule can be verified two ways: transform $f'$ directly from the table, and separately compute $sF(s) - f(0)$. They must agree — Example 1 below does exactly that.

• Dropping the initial values: $\mathcal{L}\{f'\} = sF(s)$ is wrong; the $-f(0)$ term is what carries the initial condition.
• Wrong coefficients in the second-derivative rule: it is $s^2F(s) - sf(0) - f'(0)$. The $s$ multiplies $f(0)$, never $f'(0)$.
• Confusing $\mathcal{L}\{f'(t)\}$ with $\dfrac{d}{ds}F(s)$: the first differentiates in time, the second in $s$, and the second corresponds to multiplying by $-t$.
• Transforming a constant to itself: $\mathcal{L}\{7\} = \frac{7}{s}$, not $7$. Only $\mathcal{L}\{\delta(t)\}$ equals a constant.
• Applying the rule to a discontinuous $f$: with a jump at $t=0$ the derivative includes a delta and the plain formula fails.