Higher-Order Derivatives

The derivative f'(x) of a function is itself a function — so we can differentiate it too. Doing that gives the second derivative, written f''(x). Keep going and you get the third, f'''(x), and so on.

f \;\xrightarrow{\;\frac{d}{dx}\;}\; f' \;\xrightarrow{\;\frac{d}{dx}\;}\; f'' \;\xrightarrow{\;\frac{d}{dx}\;}\; f''' \;\cdots

Each one measures the rate of change of the one before it. There is nothing new to learn: you already know how to differentiate a polynomial — you just do it again.

Differentiate, then differentiate again

Take f(x) = x^4. Apply the power rule over and over and the powers march downward, each step pulling its exponent down as a factor:

f(x) = x^4,\quad f'(x) = 4x^3,\quad f''(x) = 12x^2,\quad f'''(x) = 24x,\quad f^{(4)}(x) = 24

After four steps the x is gone and we hit a constant; one more derivative and everything is 0. Past the third derivative we switch notation to f^{(n)}(x) — the nth derivative — because the prime marks get unreadable.

See them stacked together

Below is f(x) = x^3 - 3x (bold) with its first derivative f'(x) = 3x^2 - 3 and second derivative f''(x) = 6x. Slide the tangent along the bold curve: where it is steepest, f' is largest; where f' itself is rising, f'' is positive. Each curve reads off the slope of the one above it.

Why the second derivative matters: acceleration

If s(t) is the position of a moving object at time t, then its derivative is velocity and the derivative of velocity is acceleration:

v(t) = s'(t), \qquad a(t) = v'(t) = s''(t)

So acceleration is the second derivative of position. Press play to watch a ball thrown upward: its height follows s(t) = -5t^2 + 20t, its velocity s'(t) = -20t + 20 drops steadily to zero at the top, and its acceleration s''(t) = -10 stays constant — gravity, always pulling down.

Watch it on Khan Academy