Finding acceleration (second derivative)

Differentiate $x(t) = t^3 - 6t^2 + 9t$ term by term using the power rule to get the velocity: $x'(t) = 3t^2 - 12t + 9$.

Differentiate $x'(t)$ again using the power rule: $\frac{d}{dt}[3t^2] = 6t$, $\frac{d}{dt}[-12t] = -12$, and the constant term vanishes. So $a(t) = x''(t) = 6t - 12$.

Starting from $a(t) = x''(t) = 6t - 12$, differentiate once more with respect to $t$ to obtain the jerk.

The derivative of $6t$ is $6$, and the derivative of the constant $-12$ is $0$. So $j(t) = x'''(t) = 6$, a constant function.