Second and third derivatives of an exponential function

Start with $f(x) = e^{3x}$. By the chain rule, the derivative of $e^{g(x)}$ is $e^{g(x)} \cdot g'(x)$, and here $g(x) = 3x$ so $g'(x) = 3$. This gives $f'(x) = 3e^{3x}$.

Differentiate $f'(x) = 3e^{3x}$ again using the same chain rule reasoning: the constant 3 stays, and differentiating $e^{3x}$ brings down another factor of 3, giving $f''(x) = 9e^{3x}$. Differentiating once more in the same way, $f'''(x) = 27e^{3x}$.