Single-Variable-Calculus · Unit 4 · Video 5 · Interactive Practice

Why the $n$ th Derivative of $x^n$ Is $n$ Factorial (with an Induction Proof)

IKey Formulas

Formula	Name	What it says
$Dx^n = nx^{n-1}$	Power rule	One factor comes down, the exponent drops by one
$D^k x^n = n(n-1)\cdots(n-k+1)\,x^{n-k}$	After $k$ derivatives	$k$ factors out front, $x^{n-k}$ left, for $k \le n$
$D^n x^n = n!$	The $n$ th derivative	$n! = n(n-1)\cdots 2 \cdot 1$ , a constant
$D^{n+1} x^n = 0$	One derivative more	The derivative of a constant

Key Insight: Each derivative moves one number out of the exponent and into the factor in front, so after $n$ of them the front has collected every exponent from $n$ down to $2$ and no $x$ survives. Repeated differentiation therefore does not return a function to itself in general: four derivatives send $\sin x$ back to $\sin x$ , while $n$ derivatives send $x^n$ to the constant $n!$ .

IIOne Derivative at a Time

Each derivative trades one unit of exponent for one more factor out front.

Exponent (n) Derivatives taken 0

IIIFrom a Pattern to a Theorem

Four cases are only a guess; induction makes each rung of the ladder force the next.

Step 1

Step 1 — The statement

D^n x^n = n! \quad \text{for every } n = 1, 2, 3, \ldots

Step 2 — Base case $n = 1$

Dx = 1 = 1!\ \checkmark

Step 3 — Induction hypothesis

\text{Suppose } D^n x^n = n! \text{ for one particular } n.

Step 4 — Innermost derivative first

D^{n+1} x^{n+1} = D^n\big(Dx^{n+1}\big) = D^n\big((n+1)x^{n}\big) = (n+1)\,D^n x^n

Step 5 — Apply the hypothesis

(n+1)\,D^n x^n = (n+1)\cdot n! = (n+1)!\ \checkmark

IVEvery Case in One Table

On the diagonal $k = n$ the last $x$ disappears and $n!$ is what remains.

💡 $D^{n+1}x^{n+1} = (n+1)!$ and $D^{n+1}x^{n} = 0$ sound alike but differ in one thing: whether the exponent climbs along with the number of derivatives.

VQuiz Questions

Problem 1 · Differentiate $x^5$ Five Times

Given: $f(x) = x^5$ — find $D^5 x^5$ .

✅ Correct! Five derivatives bring down

5, 4, 3, 2

and

1

, and

5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 = 5!

❌ Only one factor.

5

is the factor the first derivative brings down; four more derivatives are still to come.

❌ You stopped early.

20 = 5 \cdot 4

is the coefficient after two derivatives, when

D^2 x^5 = 20x^3

still carries an

x^3

❌ That is $6!$ . The factors run from the exponent

5

downward, so the product is

5!

, not

6!

Show solution

Apply the power rule five times, each derivative lowering the exponent by one and multiplying the front by the exponent it removed:

Dx^5 = 5x^4, \qquad D^2x^5 = 20x^3, \qquad D^3x^5 = 60x^2

D^4x^5 = 120x, \qquad D^5x^5 = 120

The last derivative is the derivative of $120x$ , which is the constant $120$ . Out front the factors have accumulated as

5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 = 5!

so $D^5x^5 = 5! = \mathbf{120}$ , with no $x$ left.

Problem 2 · Stopping Part-Way

Given: $f(x) = x^5$ — find $D^3 x^5$ , the third derivative.

✅ Correct! Three derivatives bring down

5 \cdot 4 \cdot 3 = 60

and leave the exponent at

5 - 3 = 2

❌ The coefficient is right, the exponent is not. Each of the three derivatives lowers the exponent, so it falls from

5

2

, not to

3

❌ One factor short.

20 = 5 \cdot 4

counts only two derivatives; the third brings down the exponent

3

as well, giving

60

❌ That coefficient belongs to the fifth derivative.

120 = 5!

is what you get once all five factors have come down, and by then no

x

is left at all.

Show solution

The general formula for $k$ derivatives of $x^n$ collects $k$ factors, starting at $n$ :

D^k x^n = n(n-1)\cdots(n-k+1)\,x^{n-k}

With $n = 5$ and $k = 3$ the factors are $5, 4, 3$ and the exponent left is $5 - 3 = 2$ :

D^3 x^5 = 5 \cdot 4 \cdot 3\, x^{2} = \mathbf{60x^2}

Check by hand: $Dx^5 = 5x^4$ , then $D(5x^4) = 20x^3$ , then $D(20x^3) = 60x^2\ \checkmark$

Problem 3 · Carry Out the Induction Step

Given: the induction hypothesis $D^n x^n = n!$ for one particular $n$ — carry out the step that proves the claim for $n + 1$ .

Doing the innermost differentiation first, what does $D^{n+1}x^{n+1}$ become?

Applying the hypothesis to that expression gives:

✅ Correct!

D^{n+1}x^{n+1} = (n+1)\,D^n x^n = (n+1)\cdot n! = (n+1)!

, which is exactly the claim for

n+1

— the induction is complete.

❌ Peel off one derivative first. Write

D^{n+1}x^{n+1} = D^n\big(Dx^{n+1}\big)

. The inner power rule gives

Dx^{n+1} = (n+1)x^n

, and the constant

(n+1)

comes outside, leaving

(n+1)\,D^n x^n

— one derivative was spent, so

n

of them remain.

❌ Multiply, do not add.

(n+1)\cdot n! = (n+1)\cdot n\,(n-1)\cdots 2 \cdot 1

, and that product of every whole number from

n+1

down to

1

is the definition of

(n+1)!

Show solution

Step 1 — Do the innermost derivative first. Split one derivative off the front of the operator:

D^{n+1}x^{n+1} = D^n\big(Dx^{n+1}\big)

Step 2 — Power rule on the inner derivative. $Dx^{n+1} = (n+1)x^n$ , and a constant multiple passes through $D^n$ :

D^n\big((n+1)x^{n}\big) = (n+1)\,D^n x^n

Step 3 — Use the hypothesis. The supposition says $D^n x^n = n!$ , so

D^{n+1}x^{n+1} = (n+1)\cdot n! = (n+1)\,n\,(n-1)\cdots 1 = (n+1)!

Together with the base case $Dx = 1 = 1!$ , the claim holds for every positive whole number $n$ .

Problem 4 · A Whole Polynomial

Given: $p(x) = 3x^{10} - 7x^{9} + 2x^{4}$ — find $D^{10}p(x)$ .

✅ Correct! Only the degree-

10

term survives ten derivatives, and it contributes

3 \cdot 10! = 10{,}886{,}400

❌ The constant multiple stays.

D^{10}\big(3x^{10}\big) = 3\,D^{10}x^{10} = 3 \cdot 10!

— the

3

rides along through every derivative.

❌ The $x^9$ term is already gone.

D^9x^9 = 9!

is a constant, so the tenth derivative kills it:

D^{10}x^9 = 0

❌ Not every term dies. Terms of degree below

10

vanish; the degree-

10

term reaches the constant

10!

on exactly the tenth derivative.

Show solution

Differentiation is term by term, so handle the three terms separately.

The $x^{10}$ term: ten derivatives leave the constant $10!$ , and the coefficient $3$ passes through:

D^{10}\big(3x^{10}\big) = 3 \cdot 10! = 3 \cdot 3{,}628{,}800 = 10{,}886{,}400

The $x^{9}$ term: nine derivatives already reduce it to a constant, $D^{9}x^{9} = 9!$ , and the tenth derivative of a constant is $0$ :

D^{10}\big(-7x^{9}\big) = 0

The $x^{4}$ term: it is constant after four derivatives, so $D^{10}\big(2x^4\big) = 0$ for the same reason.

D^{10}p(x) = 3 \cdot 10! = \mathbf{10{,}886{,}400}

In general $D^{k}x^{n} = 0$ whenever $k > n$ : the exponent runs out before the derivatives do.

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Why the nnnth Derivative of xnx^nxn Is nnn Factorial (with an Induction Proof)

IKey Formulas

IIOne Derivative at a Time

IIIFrom a Pattern to a Theorem

IVEvery Case in One Table

VQuiz Questions

Why the $n$ th Derivative of $x^n$ Is $n$ Factorial (with an Induction Proof)