Power rule — proof for integer, rational exponents: Worked examples

Before symbols, one reminder of what every symbol means — earn them all:

Definition The symbols we reuse everywhere

$x$ — the input, a number we are free to nudge (the horizontal position on a graph).
$x^{n}$ — " $x$ multiplied by itself, in the pattern set by $n$ ". If $n = 3$ it is $x \cdot x \cdot x$ ; if $n = - 2$ it is $\frac{1}{x \cdot x}$ ; if $n = \frac{1}{2}$ it is $x$ .
$h$ — a tiny nudge we add to $x$ . The slope is measured by comparing the height at $x + h$ with the height at $x$ , then shrinking $h$ toward $0$ (this is the Limit definition of the derivative). Whenever you see " $h \to 0$ ", picture the nudge vanishing.
$\frac{d}{d x} ()$ — "the slope of the graph of whatever is inside, at position $x$ ". A slope is rise over run: how many units up the curve climbs for one unit right. A big slope = steep hill; slope $0$ = flat ground; negative slope = downhill.

The power rule is a shortcut for that slope: drop the exponent to the front, then lower the exponent by one.

The scenario matrix

Every problem in this whole topic is one of the cells below. The examples that follow are tagged with the cell they cover.

Cell	Case class	What's tricky about it	Example
A	Positive integer exponent	none — the "easy" base case	Ex 1
B	Exponent $= 0$ / a plain constant	slope of a flat line is $0$	Ex 2
C	Negative integer exponent	answer's sign flips; undefined at $x = 0$	Ex 3
D	Rational exponent / root	fraction arithmetic; cusp at $x = 0$	Ex 4
E	"Hidden" power — needs algebra first	$3 x$ , $\frac{1}{x ^{5}}$ , $\frac{x ^{2}}{x}$ must be rewritten as $x^{n}$	Ex 5
F	Evaluate the slope at a point (sign of slope)	is the curve going up or down there?	Ex 6
G	Degenerate / undefined input	slope blows up as $x \to 0$	Ex 7
H	Real-world rate (with units)	a word problem; keep the units honest	Ex 8
I	Irrational exponent	does the shortcut still work for $x^{π}$ ?	Ex 9
J	Exam twist — looks like power rule but isn't	$2^{x}$ is exponential, not a power	Ex 10

Example 1 — Cell A: positive integer

Worked example Differentiate

$f (x) = x^{7}$ . Forecast: cover the answer and guess. Drop the $7$ , lower the exponent... what do you get?

Step 1 — identify $n$ . Here $n = 7$ , a positive whole number. Why this step? The rule $\frac{d}{d x} x^{n} = n x^{n - 1}$ needs us to name $n$ first, so we know what to drop and subtract.

Step 2 — drop the exponent to the front. That gives a factor of $7$ . Why this step? This is the first "Drop" of the mnemonic.

Step 3 — lower the exponent by one. $7 - 1 = 6$ . $f^{'} (x) = 7 x^{6} .$ Why this step? Recall the nudge $h$ : when you expand $(x + h)^{7} - x^{7}$ and divide by $h$ , every term except $7 x^{6}$ still carries a factor of $h$ , so those terms shrink to $0$ as the nudge vanishes. Only $7 x^{6}$ survives.

Valid where? Everywhere — a polynomial is smooth for all $x$ .

Verify: at $x = 1$ , $f^{'} (1) = 7$ . Slope check from first principles with a tiny $h = 0.001$ : $\frac{1.00 1 ^{7} - 1}{0.001} \approx 7.02$ — close to $7$ . ✓

Example 2 — Cell B: exponent zero / constant

Worked example Differentiate

$g (x) = x^{0}$ , and separately $c (x) = 5$ . Forecast: $x^{0}$ equals $1$ (any nonzero number to the zero power is $1$ ). The graph of $y = 1$ is a flat horizontal line. What is the slope of flat ground?

Step 1 — recognise both are constants. $x^{0} = 1$ and $5$ are both just numbers; their graphs are horizontal lines. Why this step? A horizontal line never rises, so its rise-over-run is $0$ everywhere.

Step 2 — apply the rule to $x^{0}$ . With $n = 0$ : $0 \cdot x^{- 1} = 0.$ $g^{'} (x) = 0, c^{'} (x) = 0.$ Why this step? The factor we "drop to the front" is $0$ , which zeroes the whole thing — matching the flat-line intuition.

Valid where? For $c (x) = 5$ , everywhere. For $g (x) = x^{0}$ , everywhere except $x = 0$ (since $0^{0}$ is not defined) — but the slope is $0$ at every point where it is defined.

Verify: slope of any horizontal line is $0$ . ✓ (No units, no motion — nothing changes.)

Example 3 — Cell C: negative integer

Worked example Differentiate

$f (x) = x^{- 5}$ . Forecast: the answer will have a negative number out front. Guess: drop $- 5$ , then $- 5 - 1 = ?$

Step 1 — identify $n$ . $n = - 5$ . Why this step? The rule holds for any real $n$ ; the parent note proved the negative-integer case by writing $x^{- 5} = 1/ x^{5}$ and taking a common denominator, so we may just apply the shortcut.

Step 2 — drop $n$ to the front. Factor of $- 5$ .

Step 3 — lower the exponent by one. $- 5 - 1 = - 6$ . The trap is to add: it is subtract, so it goes more negative. $f^{'} (x) = - 5 x^{- 6} = \frac{- 5}{x ^{6}} .$ Why this step? The general pattern for a negative-integer exponent is $\frac{d}{d x} x^{- k} = - k x^{- k - 1}$ (here the whole number $k = 5$ ).

Valid where? For all $x \neq = 0$ . The original function $x^{- 5} = 1/ x^{5}$ itself is undefined at $x = 0$ (you cannot divide by zero), so neither the function nor its slope exists there — the formula $- 5/ x^{6}$ correctly blows up as $x \to 0$ .

Verify: $f = 1/ x^{5}$ is positive and decreasing for $x > 0$ (bigger $x$ ⇒ smaller value), so the slope should be negative — and $- 5/ x^{6}$ is indeed negative there. ✓

Example 4 — Cell D: rational exponent / root

Worked example Differentiate

$f (x) = x^{2/3}$ . Forecast: $n = \frac{2}{3}$ . The front factor is $\frac{2}{3}$ ; the new exponent is $\frac{2}{3} - 1$ . What is $\frac{2}{3} - 1$ ?

Step 1 — identify $n = \frac{2}{3}$ (this is $3 x^{2}$ , the cube root of $x^{2}$ ). Why this step? We can't expand $x^{2/3}$ with a nudge $h$ the way we did for whole powers, so the parent note used a different route: set $y = x^{2/3}$ , cube both sides to get $y^{3} = x^{2}$ — now both exponents are whole numbers, which we already know how to differentiate. Differentiating that relation (this is Implicit differentiation, because $y$ secretly depends on $x$ ) recovers exactly the shortcut $n x^{n - 1}$ . So the same "drop and lower" works for fractions.

Step 2 — drop $\frac{2}{3}$ to the front.

Step 3 — subtract $1$ from the exponent. Write $1 = \frac{3}{3}$ , so $\frac{2}{3} - \frac{3}{3} = - \frac{1}{3}$ . $f^{'} (x) = \frac{2}{3} x^{- 1/3} = \frac{2}{3 3 x} .$ Why this step? The common mistake is fumbling $\frac{2}{3} - 1$ ; always turn the $1$ into $\frac{3}{3}$ first.

Valid where? Because the cube root is defined for negative numbers too, $f (x) = x^{2/3}$ exists for all $x$ — but the slope $\frac{2}{3 3 x}$ is undefined at $x = 0$ . There the curve has a cusp (defined above — a sharp point where the tangent flips): coming in from the left the tangent points steeply up-left, from the right steeply up-right, and the two don't agree, so no single slope exists at $x = 0$ (look at the sharp tip in the figure). For $x < 0$ the formula still applies (e.g. at $x = - 8$ , $3 - 8 = - 2$ , so $f^{'} (- 8) = \frac{2}{3 \cdot ( - 2 )} = - \frac{1}{3}$ : the curve descends on the left branch).

Verify: at $x = 8$ : $f^{'} (8) = \frac{2}{3} \cdot 8^{- 1/3} = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$ . Numerically $\frac{8.00 1 ^{2/3} - 8 ^{2/3}}{0.001} \approx 0.333$ . And at $x = - 8$ : $f^{'} (- 8) = - \frac{1}{3}$ . ✓

Example 5 — Cell E: hidden powers (rewrite first!)

Worked example Differentiate

$f (x) = \frac{x ^{2}}{x}$ . Forecast: you cannot apply the power rule to a fraction directly. First guess: can we rewrite this as a single $x^{something}$ ?

Step 1 — rewrite as one power. $x = x^{1/2}$ , and dividing subtracts exponents: $\frac{x ^{2}}{x ^{1/2}} = x^{2 - 1/2} = x^{3/2} .$ Why this step? The power rule only sees $x^{n}$ . Turning the expression into a single clean $x^{3/2}$ is the whole battle; the differentiation is then trivial.

Step 2 — apply the rule with $n = \frac{3}{2}$ . Drop $\frac{3}{2}$ ; new exponent $\frac{3}{2} - 1 = \frac{1}{2}$ . $f^{'} (x) = \frac{3}{2} x^{1/2} = \frac{3}{2} x .$ Why this step? This is where "rewrite before you differentiate" saves you from the Quotient rule, which would be slower and error-prone here.

Valid where? For $x > 0$ (the $x$ in the original needs a non-negative input, and the division needs $x \neq = 0$ ).

Verify: at $x = 4$ : $f^{'} (4) = \frac{3}{2} 4 = \frac{3}{2} \cdot 2 = 3$ . Numeric slope $\frac{f ( 4.001 ) - f ( 4 )}{0.001} \approx 3.0$ . ✓

Example 6 — Cell F: sign of the slope at a point (geometry)

Worked example For

$f (x) = x^{3}$ , find the slope at $x = - 2$ and at $x = 2$ . Is the curve going up or down at each? Forecast: $x^{3}$ is one connected curve. Guess: does it ever go downhill?

Step 1 — differentiate. $f^{'} (x) = 3 x^{2}$ (Cell A: $n = 3$ ).

Step 2 — plug in both points. $f^{'} (- 2) = 3 (- 2)^{2} = 3 \cdot 4 = 12, f^{'} (2) = 3 (2)^{2} = 12.$ Why this step? $(- 2)^{2} = + 4$ — squaring kills the sign. So the slope is $+ 12$ at both points. The tangent lines are equally steep and both point uphill.

Step 3 — read the geometry (look at the figure below: the red tangent lines both climb). $At both x = \pm 2 : slope = 12 > 0 \Rightarrow curve is rising.$ Why this step? Since $3 x^{2} \geq 0$ for all $x$ , the cubic never goes downhill — a fact you would miss without checking the sign.

Valid where? Everywhere — smooth polynomial.

Verify: $f^{'} (x) = 3 x^{2}$ is zero only at $x = 0$ and positive elsewhere; slopes at $\pm 2$ are both $12$ . ✓

Example 7 — Cell G: degenerate input (slope blows up)

Worked example For

$f (x) = x = x^{1/2}$ , what happens to the slope as $x \to 0^{+}$ ? Forecast: near $x = 0$ the $x$ curve looks almost vertical. Guess: does the slope stay finite or shoot to infinity?

Step 1 — differentiate. $n = \frac{1}{2}$ , so $f^{'} (x) = \frac{1}{2} x^{- 1/2} = \frac{1}{2 x}$ (Cell D).

Step 2 — probe the limit. As $x \to 0^{+}$ , $x \to 0$ , so $\frac{1}{2 x} \to + \infty$ . $lim_{x \to 0^{+}} f^{'} (x) = + \infty.$ Why this step? The formula stays valid, but at $x = 0$ itself the slope is undefined — the tangent is vertical.

Step 3 — the other end. As $x \to \infty$ , $\frac{1}{2 x} \to 0$ : the curve flattens out. Why this step? Covering both limiting directions is the point of this cell — steep near $0$ , flat far out (see the figure).

Valid where? For $x > 0$ . At $x = 0$ the tangent is vertical (no finite slope); $x$ is not defined for $x < 0$ .

Verify: $f^{'} (0.01) = \frac{1}{2 0.01} = \frac{1}{0.2} = 5$ (large); $f^{'} (100) = \frac{1}{2 \cdot 10} = 0.05$ (tiny). ✓

Example 8 — Cell H: real-world rate with units

Worked example A balloon's volume is

$V (r) = \frac{4}{3} π r^{3}$ (cm $^{3}$ , with radius $r$ in cm). How fast does volume grow per cm of radius when $r = 5$ cm? Forecast: "how fast volume grows per cm of radius" is a slope: $\frac{d V}{d r}$ . Guess which power of $r$ survives.

Step 1 — differentiate, treating $\frac{4}{3} π$ as a constant multiplier. With $n = 3$ : $\frac{d}{d r} r^{3} = 3 r^{2}$ . $\frac{d V}{d r} = \frac{4}{3} π \cdot 3 r^{2} = 4 π r^{2} .$ Why this step? A constant out front just rides along — the power rule acts only on the $r^{3}$ . (Notice $4 π r^{2}$ is the surface area — a lovely sanity signpost.)

Step 2 — evaluate at $r = 5$ . $\frac{d V}{d r}_{r = 5} = 4 π (25) = 100 π \approx 314.16 cm^{3} per cm .$ Why this step? The question asked for a specific radius; plug in last, after the algebra.

Valid where? For $r > 0$ (a physical radius is positive).

Verify: units are $\frac{cm ^{3}}{cm} = cm^{2}$ (an area rate — matches surface area). Numerically $\frac{V ( 5.001 ) - V ( 5 )}{0.001} \approx 314.2$ . ✓

Example 9 — Cell I: irrational exponent

Worked example Differentiate

$f (x) = x^{π}$ (yes, the exponent is the number $π \approx 3.14159$ ). Forecast: does "drop and lower" still work when the exponent isn't a fraction at all?

Step 1 — check the rule still applies. $π$ is a real number, and the power rule was proved for every real exponent — a fraction like $\frac{22}{7}$ obeys it, and $π$ is the limit of such fractions, so the shortcut carries over unchanged. Why this step? We are not allowed to expand $(x + h)^{π}$ with a nudge (no finite expansion exists for a non-whole power), so we lean on the fact that the rule extends to all real $n$ — you simply treat $π$ as "just a number".

Step 2 — drop $π$ to the front, lower the exponent by one. $f^{'} (x) = π x^{π - 1} .$ Why this step? Same mechanical move as always; the only oddity is that the new exponent $π - 1 \approx 2.14159$ is itself irrational — that is fine.

Valid where? For $x > 0$ (a general real power like $x^{π}$ is only defined for positive bases).

Verify: at $x = 1$ the exact answer is $f^{'} (1) = π \cdot 1^{π - 1} = π \approx 3.1416$ . To confirm this with the Limit definition of the derivative, use a tiny nudge so the secant slope (rise over run between two nearby points) approaches the tangent slope: $\frac{f ( 1 + h ) - f ( 1 )}{h}$ with $h = 0.001$ gives $\frac{1.00 1 ^{π} - 1 ^{π}}{0.001} = \frac{1.003146 - 1}{0.001} \approx 3.146$ . Why this step? As $h \to 0$ this secant slope converges to the true derivative $π \approx 3.1416$ , and $3.146$ is already close (a bigger $h$ would overshoot because the curve bends). ✓

Example 10 — Cell J: the exam twist (NOT the power rule)

Worked example Differentiate

$f (x) = 2^{x}$ . (Careful — is this a power rule problem?) Forecast: it is tempting to write $x \cdot 2^{x - 1}$ . Pause: which is the variable — the base or the exponent?

Step 1 — diagnose the structure. In $x^{n}$ the base is the variable, the exponent is fixed. In $2^{x}$ the base is fixed ( $2$ ) and the exponent is the variable. These are opposite creatures. Why this step? The power rule was proved for a variable base, constant exponent. It does not apply here at all — using it is a category error.

Step 2 — use the exponential rule instead (see Derivative of exponential functions): $\frac{d}{d x} a^{x} = a^{x} ln a \Rightarrow \frac{d}{d x} 2^{x} = 2^{x} ln 2.$ Why this step? This is the correct tool; $ln 2 \approx 0.693$ is the "growth constant" of base $2$ .

Valid where? Everywhere — $2^{x}$ is defined and smooth for all real $x$ .

Verify: at $x = 3$ : correct answer $2^{3} ln 2 = 8 \cdot 0.693 \approx 5.545$ ; the wrong power-rule guess $x 2^{x - 1} = 3 \cdot 4 = 12$ — different, confirming they are not the same rule. ✓

Recall Self-test (predict, then reveal)

Differentiate $\frac{1}{3 x}$ from scratch, and state where it is valid. ::: Rewrite as $x^{- 1/3}$ (Cell E). Then $n = - \frac{1}{3}$ : $f^{'} = - \frac{1}{3} x^{- 1/3 - 1} = - \frac{1}{3} x^{- 4/3} = \frac{- 1}{3 3 x ^{4}}$ . Valid for $x \neq = 0$ (undefined at $x = 0$ ). ✓

Active recall

\frac{d}{d x} x^{7}

7 x^{6}

\frac{d}{d x} x^{0}

and

\frac{d}{d x} 5

Both

0

(flat lines have zero slope).

\frac{d}{d x} x^{- 5}

, and where is it valid?

- 5 x^{- 6} = - 5/ x^{6}

; valid for

x \neq = 0

\frac{d}{d x} x^{2/3}

, and what happens at

x = 0

\frac{2}{3} x^{- 1/3}

; at

x = 0

there is a cusp, so no slope exists there.

What is a cusp?

A sharp point on a curve where the tangent flips direction, so no single slope exists there.

Before differentiating

\frac{x ^{2}}{x}

, what must you do?

Rewrite as one power:

x^{3/2}

Why is the slope of

x^{3}

never negative?

f^{'} = 3 x^{2} \geq 0

for all

x

(squares are non-negative).

What happens to

\frac{d}{d x} x

x \to 0^{+}

It goes to

+ \infty

(vertical tangent; undefined at

x = 0

\frac{d}{d x} x^{π}

π x^{π - 1}

(the rule works for every real exponent; valid for

x > 0

\frac{d V}{d r}

for

V = \frac{4}{3} π r^{3}

4 π r^{2}

(the surface area).

\frac{d}{d x} 2^{x} = x 2^{x - 1}

No —

2^{x}

is exponential; answer is

2^{x} ln 2

Connections

Power rule — proof for integer, rational exponents — the parent this page drills.
Limit definition of the derivative — the meaning of the nudge $h$ ; every numeric slope-check echoes it.
Binomial theorem — behind the integer cases (Ex 1, 6).
Implicit differentiation — why the shortcut extends to rational exponents (Ex 4).
Chain rule — needed once you nest functions inside a power.
Quotient rule — the slower alternative we avoided in Ex 5.
Derivative of exponential functions — the correct tool for Ex 10.