Derivatives of Derivative of a Function
The derivative $f'(x)$ of a function is itself a function of $x$ — it has its own value at every point, its own shape, its own behaviour. Since it is a function in its own right, we can differentiate again.
Differentiating $f'(x)$ once more produces the second derivative, written $f”(x)$ or $\dfrac{d^2f}{dx^2}$, and the process can continue: differentiating $f”(x)$ gives the third derivative $f”'(x)$, and so on.
$\qquad f'(x) = \frac{df}{dx},$
$ \qquad f”(x) = \frac{d}{dx}\left(\frac{df}{dx}\right) = \frac{d^2f}{dx^2},$
$ \qquad f”'(x) = \frac{d}{dx}\left(\frac{d^2f}{dx^2}\right) = \frac{d^3f}{dx^3}$
$\qquad \ddots \ddots \ddots \ddots \ddots \ddots \ddots$
Each derivative answers a question one level removed from the one before it. If $f(x)$ describes a quantity, $f'(x)$ describes how fast that quantity is changing, and $f”(x)$ describes how fast that **rate of change itself** is changing — whether the rate is speeding up, slowing down, or holding steady.
**Example: $f(x) = x^4$**
$\qquad f'(x) = 4x^3,$
$\qquad f”(x) = 12x^2,$
$\qquad f”'(x) = 24x,$
$\qquad f^{(4)}(x) = 24,$
$ \qquad f^{(5)}(x) = 0$
For a power function with integer exponent $x^k, k \in \mathbb{N}$, each differentiation reduces the power by one, until eventually the function becomes a constant, and one further differentiation sends it to zero — exactly as expected, since the derivative of a constant is always zero.
Example: $f(x) = \sin x$
$\qquad f'(x) = \cos x,$
$\qquad f”(x) = -\sin x,$
$\qquad f”'(x) = -\cos x,$
$\qquad f^{(4)}(x) = \sin x$
Here the pattern never settles to zero — it cycles back to the original function after four differentiations, and then repeats indefinitely.
Example: $f(x) = e^x$
$\qquad f'(x) = e^x,$
$\qquad f”(x) = e^x,$
$ \qquad f”'(x) = e^x$
Every derivative of $e^x$ is $e^x$ itself — the one function whose rate of change, and whose rate of the rate of change, and so on indefinitely, all coincide with the function itself.
Why a second derivative?
A first derivative alone says whether a quantity is changing (increasing or decreasing) at a point. It says nothing about whether that increase is accelerating or tapering off.
Two functions can both be increasing at a point yet behave very differently just beyond it, one continuing to climb ever faster, the other beginning to level out.
The second derivative is what distinguishes these two situations: it measures whether the rate of change is itself growing or shrinking, which is exactly the information the first derivative alone cannot supply.
Everyday illustration – The motion of a particle
If $f(t)$ describes the position of a particle at time $t$, the first derivative $f'(t)$ is its velocity — how fast its position is changing.
The second derivative $f”(t)$ is its acceleration — how fast that velocity itself is changing.
A particle can be moving fast yet not accelerating at all (constant velocity), or moving slowly yet accelerating sharply (just starting to speed up) — position and velocity alone cannot tell these apart; only the second derivative can.
Recall “Rapidity”
This same idea — a quantity, its rate of change, and the rate of change of that rate — recurs in any setting where it matters not just whether something is moving, but how its motion is itself evolving.
Same velocity, different acceleration — a worked illustration
Consider three particles, each with a different position function, all examined at the same instant $t = 1$.
Particle A: $f(t) = t^2 + 4t – 2$
$ \qquad f'(t) = 2t + 4 \implies f'(1) = 6,$
$ \qquad f”(t) = 2 \implies f”(1) = 2$
Particle B: $f(t) = t^3 + t + 4$
$ \qquad f'(t) = 3t^2 + 1 \implies f'(1) = 4,$
$\qquad f”(t) = 6t \implies f”(1) = 6$
Particle C: $f(t) = 6t$
$ \qquad f'(t) = 6 \implies f'(1) = 6,$
$ \qquad f”(t) = 0 \implies f”(1) = 0$
Particle A and Particle C have exactly the same velocity at $t=1$, namely $6$ units per second — at that single instant, anyone only measuring speed would find them indistinguishable.
Yet Particle A is accelerating ($f”(1) = 2$, speeding up), while Particle C is moving at constant velocity ($f”(1) = 0$, neither speeding up nor slowing down) — Particle C, in fact, is exactly the linear function, whose own rate of change never changes at all.
Particle B, has a lower velocity than A or C at this instant ($f'(1) = 4$), but is accelerating the fastest of the three ($f”(1) = 6$) — a moment later, it may well have overtaken both, even though right now it is the slowest.
This is precisely why velocity alone, at a single instant, is not enough to describe what a moving quantity is about to do next — the same value of $f’$ can sit on top of entirely different values of $f”$, and only the second derivative reveals which one.
A second illustration — rapidity using exponential and logarithmic growth
The previous three particles were all polynomial, where acceleration itself stays simple — constant, or changing at a steady rate.
A sharper illustration of “rapidity” comes from comparing an exponential and a logarithmic motion against a linear one, since these are exactly the primitives whose own rate of change behaves in fundamentally different ways, as already seen earlier.
Particle D: $f(t) = e^{2t}$
$\qquad f'(t) = 2e^{2t} \implies f'(1) = 2e^{2} \approx 14.78,$
$\qquad f”(t) = 4e^{2t} \implies f”(1) = 4e^{2} \approx 29.56$
Particle E: $f(t) = 15t$
$\qquad f'(t) = 15 \implies f'(1) = 15,$
$ \qquad f”(t) = 0 \implies f”(1) = 0$
Particle F: $f(t) = 10\ln(t) + 15t$, for $t > 0$
$\qquad f'(t) = \frac{10}{t} + 15 \implies f'(1) = 10 + 15 = 25,$
$ \qquad f”(t) = -\frac{10}{t^2} \implies f”(1) = -10$$
At $t=1$, Particle D and Particle E have nearly the same velocity — $14.78 \approx 15$ units per second, close enough that a single speed reading would not tell them apart.
But their accelerations could not be more different: Particle D’s acceleration is itself almost twice its velocity ($29.56$), and it will only keep compounding, since the exponential’s rate of change is always proportional to how large it already is.
Particle E’s acceleration is exactly zero — its velocity at $t=1$ is the velocity it will keep forever.
Particle F’s velocity at $t=1$ is the highest of all, $25$ units per second — and yet its acceleration is negative, $-10$. It is momentarily the fastest-moving particle of the group, while already slowing down.
A moment later, having started ahead of D and E, it will fall behind both — D because exponential growth eventually outruns everything, and even E, the steady linear motion, simply by virtue of never decelerating at all.
The same instantaneous velocity, or even the highest velocity in a group, says nothing about which direction things are headed next — only the second derivative, tracking the primitive function’s own characteristic rate of change, reveals that.