Higher-Order Derivatives, calculus lesson

A derivative is itself a function, so you can ask how fast it changes. For a car, velocity is the rate of change of position, and acceleration is the rate of change of velocity. When you press the brake harder, the acceleration itself changes, and that rate has a name too. Repeating differentiation gives a whole ladder of functions, each describing how the one before it changes.

Key terms

Second derivative

Definition Suppose $f$ is differentiable on an open interval containing $a$ . If the derivative $f'$ is itself differentiable at $a$ , the second derivative of $f$ at $a$ is

f''(a) = (f')'(a) = \lim_{h \to 0} \frac{f'(a + h) - f'(a)}{h}.

Then $f$ is called twice differentiable at $a$ .

In plain words The second derivative measures how fast the slope is changing. A positive $f''$ means the slope is going up as you move right.

Example For $f(x) = x^4 - 3x^2$ the first derivative is $f'(x) = 4x^3 - 6x$ , and differentiating again gives $f''(x) = 12x^2 - 6$ . So $f''(1) = 6$ and $f''(0) = -6$ .

Higher derivatives defined recursively

Definition Set $f^{(0)} = f$ . For $n \ge 1$ , if $f^{(n-1)}$ is defined on an open interval containing $a$ and is differentiable at $a$ , the $n$ th derivative of $f$ at $a$ is

f^{(n)}(a) = \left(f^{(n-1)}\right)'(a).

A function is $n$ times differentiable on an open interval $I$ if $f^{(n)}(x)$ exists for every $x$ in $I$ .

In plain words To get the $n$ th derivative, differentiate $n$ times in a row, one step at a time. Often a pattern shows up after a few steps and lets you jump ahead.

Example The derivatives of $\sin x$ repeat with period 4.

$n$	$0$	$1$	$2$	$3$	$4$
$f^{(n)}(x)$	$\sin x$	$\cos x$	$-\sin x$	$-\cos x$	$\sin x$

Since $27 = 4 \cdot 6 + 3$ , the 27th derivative of $\sin x$ is $-\cos x$ . For a polynomial the ladder ends. If $f(x) = x^5$ then $f^{(5)}(x) = 5! = 120$ and $f^{(6)}(x) = 0$ .

Notation

Definition For $y = f(x)$ the common notations for higher derivatives are

f''(x) = y'' = \frac{d^2 y}{dx^2} = D^2 f(x), \qquad f'''(x) = y''' = \frac{d^3 y}{dx^3}, \qquad f^{(n)}(x) = y^{(n)} = \frac{d^n y}{dx^n}.

Primes are used up to the third derivative. From the fourth on, the order goes in parentheses, $f^{(4)}$ , to avoid confusion with a power $f^4$ .

In plain words The Leibniz form $\frac{d^2 y}{dx^2}$ reads as “differentiate twice with respect to $x$ ”. The 2 sits on $d$ in the top and on $x$ in the bottom because it counts the steps, not a power of $x$ .

Example For $y = \frac{1}{x}$ with $x \neq 0$ ,

\frac{dy}{dx} = -x^{-2}, \qquad \frac{d^2 y}{dx^2} = 2x^{-3}, \qquad \frac{d^3 y}{dx^3} = -6x^{-4},

and in general, for every $n \ge 1$ ,

\frac{d^n y}{dx^n} = (-1)^n n! \, x^{-(n+1)}.

Acceleration and jerk

Definition If $s(t)$ is the position of an object moving along a line at time $t$ , then its velocity is $v(t) = s'(t)$ , its acceleration is

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

and its jerk is $j(t) = a'(t) = s'''(t)$ , whenever these derivatives exist.

In plain words Acceleration tells you how quickly the velocity changes. Jerk tells you how quickly the acceleration changes, which is what you feel as a sudden lurch in a train or an elevator.

Example Let $s(t) = t^3 - 6t^2 + 9t$ meters after $t$ seconds. Then $v(t) = 3t^2 - 12t + 9$ , $a(t) = 6t - 12$ and $j(t) = 6$ . At $t = 1$ the object is momentarily at rest, $v(1) = 0$ , but its acceleration is $a(1) = -6$ meters per second squared, so it is about to move backwards.

Higher derivatives of products

Definition If $f$ and $g$ are $n$ times differentiable on an open interval $I$ , then so is $fg$ , and on $I$

(fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(k)} g^{(n-k)}.

This is the general Leibniz rule. For $n = 2$ it reads

(fg)'' = f''g + 2f'g' + fg''.

In plain words The pattern is the same as in the binomial expansion of $(a + b)^n$ , with powers replaced by numbers of derivatives. It is proved by applying the product rule repeatedly.

Example Take $f(x) = x^2$ and $g(x) = e^x$ with $n = 3$ . Since $f' = 2x$ , $f'' = 2$ and $f''' = 0$ , while every derivative of $g$ is $e^x$ ,

(x^2 e^x)''' = x^2 e^x + 3 \cdot 2x e^x + 3 \cdot 2 e^x + 0 = (x^2 + 6x + 6)e^x.

Smoothness classes

Definition Let $I$ be an open interval. The class $C^0(I)$ is the set of functions continuous on $I$ . For $k \ge 1$ , $C^k(I)$ is the set of functions that are $k$ times differentiable on $I$ with $f^{(k)}$ continuous on $I$ . The class $C^\infty(I)$ is the set of functions with derivatives of every order on $I$ . These classes are nested,

C^\infty(I) \subset \cdots \subset C^2(I) \subset C^1(I) \subset C^0(I).

In plain words The class number says how many times you can differentiate and still get a continuous function. Polynomials, $e^x$ , $\sin x$ and $\cos x$ are in $C^\infty(\mathbb{R})$ .

Example Let $f(x) = x \lvert x \rvert$ . For $x > 0$ , $f(x) = x^2$ and $f'(x) = 2x$ . For $x < 0$ , $f(x) = -x^2$ and $f'(x) = -2x$ . At $0$ the difference quotient is

\frac{f(h) - f(0)}{h} = \frac{h \lvert h \rvert}{h} = \lvert h \rvert \to 0 \quad \text{as } h \to 0,

so $f'(0) = 0$ . Hence $f'(x) = 2\lvert x \rvert$ for every $x$ , which is continuous but has a corner at $0$ , so $f''(0)$ does not exist. Therefore $f$ is in $C^1(\mathbb{R})$ but not in $C^2(\mathbb{R})$ .

Check yourself

Find $f''(x)$ for $f(x) = \ln x$ , $x > 0$ .

Show the answer

The first derivative is $f'(x) = \frac{1}{x} = x^{-1}$ , so $f''(x) = -x^{-2} = -\frac{1}{x^2}$ .

What is the 100th derivative of $e^{2x}$ ?

Show the answer

Each differentiation multiplies by 2, so

\frac{d^{100}}{dx^{100}} e^{2x} = 2^{100} e^{2x}.

A particle has position $s(t) = t^4$ . Find its acceleration and jerk at $t = 2$ .

Show the answer

Differentiating gives $s' = 4t^3$ , then $s'' = 12t^2$ and $s''' = 24t$ , so $a(2) = 48$ and $j(2) = 48$ .

Higher-Order Derivatives

Theory, part 55

Practice, part 56

Key terms

Second derivative

Higher derivatives defined recursively

Notation

Acceleration and jerk

Higher derivatives of products

Smoothness classes

Check yourself

How this lesson connects

Builds on

Used later in