Single-Variable-Calculus · Unit 4 · Video 4 · Interactive Practice

Higher Derivatives and Their Notation: Primes, D, and d²u/dx²

IKey Formulas

Formula	Name	What it says
$u'' = \dfrac{d}{dx}\!\left(\dfrac{du}{dx}\right) = D^{2}u = \dfrac{d^{2}u}{dx^{2}}$	Second derivative	Differentiate, then differentiate the result
$f^{(n)}(x) = D^{n}f = \dfrac{d^{n}f}{dx^{n}}$	$n$ -th derivative, three notations	$n$ counts the differentiations in all three
$D = \dfrac{d}{dx}$	The operator	Function in, function out
$u = \sin x \;\Longrightarrow\; u' = \cos x,\ u'' = -\sin x,\ u''' = -\cos x,\ u^{(4)} = \sin x$	The sine cycle	Four differentiations return the function

Key Insight: The chain $u'' = \dfrac{d}{dx}\!\left(\dfrac{du}{dx}\right) = \left(\dfrac{d}{dx}\right)^{2}u = D^{2}u = \dfrac{d^{2}u}{dx^{2}}$ carries no mathematical content beyond this: five ways of writing, one derivative. In the last form the denominator is $(dx)^{2}$ — $x$ is never squared.

IIThe Sine Ring

Differentiate $\sin x$ four times and the curve lands back on the one you started from.

Differentiations (n) 0.00

💡 The ring belongs to sine and cosine alone: a polynomial's derivatives fall in degree until they reach zero and stay there.

IIIOne Object, Three Spellings

Each application of $D$ takes one function in and hands a new function out.

Function

💡 A fifth application separates the two for good: $D^{5}x^{4} = 0$ , while $D^{5}\sin x = \cos x$ and the ring turns again.

IVReading the Denominator

The twos in $\dfrac{d^{2}u}{dx^{2}}$ count differentiations; the denominator is $(dx)^{2}$ , never $d(x^{2})$ .

Step 1

Step 1 — Differentiate twice

u'' = \frac{d}{dx}\!\left(\frac{du}{dx}\right)

Step 2 — Collect the numerator

d \cdot d \cdot u = d^{2}u

Step 3 — Collect the denominator

dx \cdot dx = (dx)^{2}, \ \text{written } dx^{2}

Step 4 — What the symbol says

\frac{d^{2}u}{dx^{2}} = \frac{d^{2}u}{(dx)^{2}} \ \neq \ \frac{d^{2}u}{d(x^{2})}

VQuiz Questions

Problem 1 · Three Steps Round the Ring

Given: $u = \cos x$ — find $u'''$ .

✅ Correct!

\cos x \to -\sin x \to -\cos x \to \sin x

: three prime marks, three differentiations.

❌ That is $u''$ . One differentiation short —

\dfrac{d}{dx}(-\cos x) = \sin x

, because the derivative of

\cos x

-\sin x

and the two minus signs cancel.

❌ That is $u'$ . Only one differentiation has been done;

u'''

asks for three.

❌ That is $u^{(4)}$ . One differentiation too many — the ring closes on the fourth step, and

u'''

is the step before it.

❌ Not quite. Differentiate one step at a time, using

\dfrac{d}{dx}\sin x = \cos x

and

\dfrac{d}{dx}\cos x = -\sin x

Show solution

Each step uses one of the two trig derivatives, with the constant $-1$ coming out in front when it is there:

u = \cos x

u' = -\sin x

u'' = -\cos x

u''' = -(-\sin x) = \sin x

At the third step the derivative of $-\cos x$ is $-(-\sin x)$ , and that new minus sign cancels the one already there. One more differentiation would give $u^{(4)} = \cos x$ : cosine sits on the same ring as sine and returns to itself after four steps too.

Problem 2 · What the Symbol Stands For

Given: the symbol $\dfrac{d^{2}u}{dx^{2}}$ — choose the expression it is shorthand for.

✅ Correct! Both twos are counts of differentiations:

d \cdot d = d^{2}

on top,

dx \cdot dx = (dx)^{2}

underneath.

❌ Squaring is a different operation. For

u = \sin x

\left(\dfrac{du}{dx}\right)^{2} = \cos^{2}x

, while

\dfrac{d^{2}u}{dx^{2}} = -\sin x

❌ That is the reading the notation warns against. The denominator is the whole symbol

dx

, squared — no

x^{2}

appears anywhere, and nothing is being done to

x^{2}

❌ The exponent is not a multiplier. For

u = \sin x

it would give

2\cos x

, whereas the second derivative is

-\sin x

Show solution

The second derivative is the first derivative differentiated, so it is $\dfrac{d}{dx}$ applied to $\dfrac{du}{dx}$ :

u'' = \frac{d}{dx}\!\left(\frac{du}{dx}\right) = \left(\frac{d}{dx}\right)^{2}u = D^{2}u = \frac{d^{2}u}{dx^{2}}

The last form is a bookkeeping abbreviation of the one before it: the two $d$ 's of the two differentiations are collected as $d^{2}$ in the numerator, and the two $dx$ 's as $(dx)^{2}$ in the denominator, which is written $dx^{2}$ for short.

Check with $u = \sin x$ : $\dfrac{du}{dx} = \cos x$ and $\dfrac{d^{2}u}{dx^{2}} = -\sin x$ . Neither $\cos^{2}x$ nor $2\cos x$ is that.

Problem 3 · Six Times Round

Given: $u = \sin x$ — find $D^{6}u$ .

✅ Correct!

D^{4}u = \sin x

puts you back at the start, so

D^{6}u = D^{2}\!\left(D^{4}u\right) = D^{2}\sin x = -\sin x

❌ That is $D^{4}u$ or $D^{8}u$ . The ring returns to

\sin x

only after a whole number of full turns, and

6

is not a multiple of

4

❌ That is $D^{5}u$ . Five steps round the ring, one short of six.

❌ That is $D^{7}u$ . Seven steps round the ring, one past six.

Show solution

Count the steps round the ring, four at a time:

D^{4}u = \sin x \qquad\text{(back to the start)}

D^{5}u = \cos x

D^{6}u = -\sin x

In general, only the remainder of $n$ on division by $4$ matters: remainder $0$ gives $\sin x$ , $1$ gives $\cos x$ , $2$ gives $-\sin x$ , $3$ gives $-\cos x$ . Here $6 = 4 + 2$ , so $D^{6}u$ is two steps past the start, which is $-\sin x$ .

The same answer in the other two notations: $u^{(6)}(x) = -\sin x$ and $\dfrac{d^{6}u}{dx^{6}} = -\sin x$ .

Problem 4 · A Function That Does Not Come Back

Given: $f(x) = x^{4}$ — find $D^{2}f$ and $\dfrac{d^{4}f}{dx^{4}}$ .

What is $D^{2}f$ ?

And what is $\dfrac{d^{4}f}{dx^{4}}$ ?

✅ Correct!

x^{4} \to 4x^{3} \to 12x^{2} \to 24x \to 24

: the fourth derivative is the constant

24

, and no trace of

x^{4}

comes back.

❌ That is $(Df)^{2} = (4x^{3})^{2}$ . The exponent on

D

counts differentiations, not powers:

D^{2}f

means differentiate

4x^{3}

again.

❌ Check the count.

Df = 4x^{3}

, and one more differentiation gives

D^{2}f = 12x^{2}

❌ Check the count. The four derivatives in order are

4x^{3}

12x^{2}

24x

24

— the constant arrives on the fourth step, and

0

only on the fifth.

Show solution

Differentiate four times, each step dropping the degree by one:

Df = 4x^{3}, \qquad D^{2}f = 12x^{2}, \qquad D^{3}f = 24x, \qquad D^{4}f = 24

So $D^{2}f = 12x^{2}$ and $\dfrac{d^{4}f}{dx^{4}} = 24 = 4!$ , a constant. A fifth differentiation gives $0$ , and every one after it gives $0$ as well.

This is the contrast with $\sin x$ : there the fourth derivative is the original function again, here it is a number. The returning behaviour is special to the sines and cosines, not a general fact about fourth derivatives.

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