Higher Derivatives

Once we have the derivative f' of a function $f$ , we can differentiate again to get the second derivative f'', and continue to get derivatives of any order. Higher derivatives measure how rapidly the rate of change itself is changing.

Notation	Meaning
f''(x), y'', $\frac{d^{2} y}{d x^{2}}$	Second derivative
f'''(x), y''', $\frac{d^{3} y}{d x^{3}}$	Third derivative
$f^{(n)} (x)$ , $y^{(n)}$ , $\frac{d^{n} y}{d x^{n}}$	$n$ -th derivative

Motivation: Velocity and Acceleration

If $s (t)$ is the position of an object moving on a straight line, then:

Velocity v(t)=s'(t) is the first derivative of position.
Acceleration a(t)=v'(t)=s''(t) is the second derivative of position.

Acceleration appears in Newton's second law $F = m a$ .

Definition of Higher Derivatives

If f' exists, we can take its derivative to obtain the second derivative of $f$ :

f''=(f')'

The second derivative is also written y'' or $\frac{d^{2} y}{d x^{2}}$ .

Continuing, the third derivative is

f'''=(f')'',\qquad y''',\qquad\frac{d^3y}{dx^3}

The fourth derivative is

f^{(4)}=(f''')',\qquad y^{(4)},\qquad\frac{d^4y}{dx^4}

In general, the $n$ -th derivative:

y^{(n)} = f^{(n)} (x) = \frac{d^{n} y}{d x^{n}} = \frac{d^{n} f}{d x^{n}}

If $f^{(n)} (x_{0})$ exists, then $f$ is said to be $n$ times differentiable at $x_{0}$ .

Example. For $f (x) = x^{3} - 5 x^{2} + 3 x - 1$ :

f'(x)=3x^2-10x+3,\qquad f''(x)=6x-10

Worked Examples

If $y = \sin x$ , find $y^{(4)}$ .

Solution

\begin{aligned} y&=\sin x\\ y'&=\cos x\\ y''&=-\sin x\\ y'''&=-\cos x\\ y^{(4)}&=\sin x \end{aligned} Therefore

y^{(4)} = y = \sin x

. The derivatives of

\sin x

cycle with period 4.

For $f (x) = - x^{5} + 3 x^{3} + 2 x^{2} - 1$ , find f'(x),f''(x),f'''(x), and $f^{(4)} (x)$ .

Solution

\begin{aligned} f'(x)&=-5x^4+9x^2+4x\\ f''(x)&=-20x^3+18x+4\\ f'''(x)&=-60x^2+18\\ f^{(4)}(x)&=-120x \end{aligned}

Determine $a, b$ , and $c$ such that f''(x) exists everywhere if

f(x)=\begin{cases}x^3 & \text{when }x\leq1\\ax^2+bx+c & \text{when }x>1\end{cases}

Solution

The second derivatives of each piece exist everywhere. The only issue is at

x = 1

where the formula changes. We need three conditions for f''(1) to exist: **(1) Continuity at

x = 1

:** f(1)=1^3=a\cdot1+b\cdot1+c\quad\Rightarrow\quad a+b+c=1\tag{i} **(2) First derivative exists at

x = 1

** (i.e., f'_-(1)=f'_+(1)): 3x^2\big|_{x=1}=2ax+b\big|_{x=1}\quad\Rightarrow\quad 3=2a+b\tag{ii} **(3) Second derivative exists at

x = 1

** (i.e., f''_-(1)=f''_+(1)): 6x\big|_{x=1}=2a\big|_{x=1}\quad\Rightarrow\quad 6=2a\tag{iii} From (iii):

a = 3

. From (ii):

b = 3 - 6 = - 3

. From (i):

c = 1 - 3 + 3 = 1

. Therefore

a = 3

b = - 3

c = 1

Differentiability Does Not Imply Differentiability of the Derivative

A function can be differentiable (have f' everywhere) without f' being differentiable (i.e., f'' need not exist everywhere).

Let

f(x)=\begin{cases}x^2\sin\dfrac{1}{x} & x\neq0\\0 & x=0\end{cases}

Does f'(0) exist? Is f'(x) continuous at $x = 0$ ?

Solution

**Finding f'(0) from the definition:** f'(0)=\lim_{\Delta x\to0}\frac{(\Delta x)^2\sin\frac{1}{\Delta x}}{\Delta x}=\lim_{\Delta x\to0}\Delta x\sin\frac{1}{\Delta x}=0 (since

| Δ x \sin (1 / Δ x) | \leq | Δ x | \to 0

) **Finding f'(x) for

x \neq 0

by differentiation rules:** f'(x)=2x\sin\frac{1}{x}+x^2\cdot\left(-\frac{1}{x^2}\right)\cos\frac{1}{x}=2x\sin\frac{1}{x}-\cos\frac{1}{x}\quad(x\neq0) So f'(x)=\begin{cases}2x\sin\dfrac{1}{x}-\cos\dfrac{1}{x} & x\neq0\\0 & x=0\end{cases} **Is f' continuous at

x = 0

?** \lim_{x\to0}f'(x)=\lim_{x\to0}\left(2x\sin\frac{1}{x}-\cos\frac{1}{x}\right) The term

\cos (1 / x)

oscillates endlessly as

x \to 0

and has no limit. Therefore \lim_{x\to0}f'(x) does not exist, and f' is **not continuous** at

x = 0

. Consequently f''(0) does not exist.

The above example shows: from the existence of f'(x_0), we cannot infer the existence of f''(x_0).

Frequently Asked Questions

What is the physical meaning of the second derivative?

In kinematics, the second derivative of position is acceleration. More generally, f''(x) measures how rapidly the slope of the graph of

f

is changing, which is related to the "bending" (concavity) of the curve.

For a polynomial of degree

n

, what happens to derivatives beyond order

n

They all equal zero. The

n

-th derivative of a degree-

n

polynomial

a_{n} x^{n} + \dots + a_{0}

is the constant

n! a_{n}

, and all higher derivatives are zero.

Can a differentiable function have a discontinuous derivative?

Yes. The example

f (x) = x^{2} \sin (1 / x)

(extended by

f (0) = 0

) is differentiable everywhere, but f' is not continuous at

x = 0

. This shows that differentiability is weaker than having a continuous derivative.

Motivation: Velocity and Acceleration

Definition of Higher Derivatives

Worked Examples

Differentiability Does Not Imply Differentiability of the Derivative

Frequently Asked Questions

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