Power rule — proof for integer, rational exponents: Exercises

The one rule under test everywhere below: $\frac{d}{d x} x^{n} = n x^{n - 1} .$

The two engines behind it (from the parent note): the Limit definition of the derivative and, for whole-number powers, the Binomial theorem.

Level 1 — Recognition

You just apply "drop and drop": slide $n$ to the front, then subtract one from the exponent.

Q1. Differentiate $f (x) = x^{7}$ .

Q2. Differentiate $g (x) = x^{- 4}$ .

Q3. Differentiate $h (x) = 3 x$ (that is, $x^{1/3}$ ).

Recall Solution — Q1, Q2, Q3

Q1. Here $n = 7$ . WHAT: apply $n x^{n - 1}$ . WHY: $x^{7}$ is a variable base with a constant exponent — exactly the shape the power rule handles. $f^{'} (x) = 7 x^{7 - 1} = 7 x^{6} .$

Q2. Here $n = - 4$ (a negative integer — Stage 3 of the proof covers this). $g^{'} (x) = - 4 x^{- 4 - 1} = - 4 x^{- 5} = - \frac{4}{x ^{5}} .$ WHY the extra minus: subtracting one from a negative exponent makes it more negative.

Q3. Rewrite the root as a power first: $3 x = x^{1/3}$ , so $n = \frac{1}{3}$ . $h^{'} (x) = \frac{1}{3} x^{1/3 - 1} = \frac{1}{3} x^{- 2/3} = \frac{1}{3 x ^{2/3}} = \frac{1}{3 3 x ^{2}} .$ WHY rewrite: the power rule needs an exponent, not a root symbol — same number, cleaner form.

Level 2 — Application

Now the exponents are messier, and you combine the power rule with the fact that constants multiply through and sums differentiate term-by-term.

Q4. Differentiate $f (x) = 5 x^{3} - \frac{2}{x} + 6 x$ .

Q5. Differentiate $g (x) = \frac{x ^{2} + 3 x}{x}$ . (Hint: split the fraction first.)

Q6. Find the slope of $y = x^{5/2}$ at $x = 4$ .

Recall Solution — Q4, Q5, Q6

Q4. WHAT: rewrite each piece as a clean power. $\frac{2}{x} = 2 x^{- 1}$ and $6 x = 6 x^{1/2}$ . Now differentiate term by term (a constant multiple just rides along): $f^{'} (x) = 5 \cdot 3 x^{2} - 2 \cdot (- 1) x^{- 2} + 6 \cdot \frac{1}{2} x^{- 1/2} .$ $f^{'} (x) = 15 x^{2} + \frac{2}{x ^{2}} + \frac{3}{x} .$ WHY the middle term turned positive: $- 2 \cdot (- 1) = + 2$ — two minus signs.

Q5. WHAT: split before differentiating (you cannot power-rule a quotient directly here). $g (x) = \frac{x ^{2}}{x ^{1/2}} + \frac{3 x}{x ^{1/2}} = x^{3/2} + 3 x^{1/2} .$ WHY split: dividing powers of the same base subtracts exponents — turning one messy fraction into two simple powers. Now differentiate: $g^{'} (x) = \frac{3}{2} x^{1/2} + 3 \cdot \frac{1}{2} x^{- 1/2} = \frac{3}{2} x + \frac{3}{2 x} .$

Q6. WHAT: differentiate, then plug in. $n = \frac{5}{2}$ : $y^{'} = \frac{5}{2} x^{3/2} .$ At $x = 4$ : $4^{3/2} = (4)^{3} = 2^{3} = 8$ , so $y^{'} (4) = \frac{5}{2} \cdot 8 = 20.$ WHY $4^{3/2} = 8$ : the exponent $3/2$ means "cube the square root" — take $4 = 2$ , then cube.

Level 3 — Analysis

Here you reason about the rule, not just crank it — reversing it, using it geometrically, and reading off where a derivative behaves strangely.

Q7. A curve has slope $y^{'} = 12 x^{3}$ everywhere and passes through the origin's height rule $y = x^{4}$ -type family. Which power function $y = x^{n}$ has derivative exactly $12 x^{3}$ ? (Find $n$ and the leading coefficient if $y = c x^{n}$ .)

Q8. Find the equation of the tangent line to $y = x$ at the point where $x = 9$ .

Q9. For $f (x) = x^{1/3}$ , explain — using the derivative — why the graph has a vertical tangent at $x = 0$ . (See the step figure.)

Recall Solution — Q7, Q8, Q9

Q7. WHAT: run the power rule backward. If $y = c x^{n}$ then $y^{'} = c n x^{n - 1}$ . We need this to equal $12 x^{3}$ . WHY match pieces: two power functions are equal only if their exponents match AND their coefficients match.

Exponent: $n - 1 = 3 \Rightarrow n = 4$ .
Coefficient: $c n = 12 \Rightarrow c \cdot 4 = 12 \Rightarrow c = 3$ .

So $y = 3 x^{4}$ (check: $y^{'} = 3 \cdot 4 x^{3} = 12 x^{3}$ ✓).

Q8. WHAT: get the slope, get the point, assemble the line $y - y_{0} = m (x - x_{0})$ .

Point: $y (9) = 9 = 3$ . So $(x_{0}, y_{0}) = (9, 3)$ .
Slope: $y^{'} = \frac{1}{2 x}$ , so $m = \frac{1}{2 9} = \frac{1}{6}$ . $y - 3 = \frac{1}{6} (x - 9) ⟹ y = \frac{1}{6} x + \frac{3}{2} .$ WHY point-slope: it is the fastest way to build a line from one point and one slope.

Q9. WHAT: compute $f^{'} (x)$ and watch it as $x \to 0$ . $f^{'} (x) = \frac{1}{3} x^{- 2/3} = \frac{1}{3 x ^{2/3}} .$ WHY it blows up: $x^{2/3}$ is a positive number that shrinks to $0$ as $x \to 0$ , so $1/ (3 x^{2/3})$ grows without bound: $f^{'} (x) \to + \infty$ . An infinite slope means the tangent line stands straight up — vertical. Both sides ( $x \to 0^{+}$ and $x \to 0^{-}$ ) give $x^{2/3} > 0$ , so the slope rockets to $+ \infty$ from both sides. See the near-vertical tangent in the figure below.

Figure — Power rule — proof for integer, rational exponents

Level 4 — Synthesis

Now you rebuild the rule itself — proving special cases from the Limit definition of the derivative and stitching in the Chain rule / Implicit differentiation.

Q10. Prove from the limit definition (Stage-1 style) that $\frac{d}{d x} x^{3} = 3 x^{2}$ , showing every cancellation.

Q11. Prove from first principles that $\frac{d}{d x} x^{- 1} = - x^{- 2}$ using the common-denominator trick (Stage 3).

Q12. Use implicit differentiation (Stage 4) to prove $\frac{d}{d x} x^{2/3} = \frac{2}{3} x^{- 1/3}$ .

Recall Solution — Q10, Q11, Q12

Q10. WHAT: write the limit-definition quotient and expand $(x + h)^{3}$ by the Binomial theorem. $f^{'} (x) = lim_{h \to 0} \frac{( x + h ) ^{3} - x ^{3}}{h} .$ Expand: $(x + h)^{3} = x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3}$ . Subtract $x^{3}$ (the leading term cancels): $(x + h)^{3} - x^{3} = 3 x^{2} h + 3 x h^{2} + h^{3} .$ WHY divide by $h$ : every surviving term carries at least one $h$ , so dividing is legal and exposes which term has no $h$ left: $\frac{( x + h ) ^{3} - x ^{3}}{h} = 3 x^{2} + 3 x h + h^{2} .$ Send $h \to 0$ : the last two terms each contain $h$ and vanish. $f^{'} (x) = 3 x^{2} . ✓$

Q11. WHAT: $f (x) = x^{- 1} = \frac{1}{x}$ . Form the difference quotient and combine over a common denominator.

= \lim_{h\to0}\frac{1}{h}\cdot\frac{x - (x+h)}{(x+h)x}.$$ WHY common denominator: it collapses two fractions into one whose top simplifies. The top is $x-(x+h) = -h$: $$= \lim_{h\to0}\frac{-h}{h\,(x+h)x} = \lim_{h\to0}\frac{-1}{(x+h)x}.$$ WHY the $h$ cancels: it was the only thing preventing us from setting $h=0$. Now let $h\to0$: $$f'(x) = \frac{-1}{x\cdot x} = -\frac{1}{x^2} = -x^{-2}. \checkmark$$ **Q12.** WHAT: set $y = x^{2/3}$ and cube both sides to clear the fraction (here $p=2,\,q=3$): $$y^3 = x^2.$$ WHY cube: it turns a fractional exponent into two *integer* powers we already proved. Differentiate both sides with respect to $x$; the left needs the [[Chain rule]] because $y$ depends on $x$: $$3y^2\,\frac{dy}{dx} = 2x.$$ Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{2x}{3y^2}.$$ Substitute $y = x^{2/3}$, so $y^2 = x^{4/3}$: $$\frac{dy}{dx} = \frac{2x}{3\,x^{4/3}} = \tfrac23\,x^{1 - 4/3} = \tfrac23\,x^{-1/3}. \checkmark$$

Level 5 — Mastery

Boundary cases, wrong-rule detection, and a limit-of-the-rule sanity check — where deep understanding separates from mechanical use.

Q13. Two look-alikes: differentiate (a) $y = x^{e}$ and (b) $y = e^{x}$ . Which one uses the power rule, which does not, and why? Give both derivatives.

Q14. Zero/degenerate case. Consider $f (x) = x^{0}$ (defined for $x \neq = 0$ ). Compute $f^{'} (x)$ two ways: (i) directly, and (ii) by the power-rule formula $n x^{n - 1}$ with $n = 0$ . Confirm they agree, and explain what the formula's factor of $n$ is "protecting" us from.

Recall Solution — Q13, Q14

Q13. The deciding question: is the exponent constant with a variable base, or is the base constant with a variable exponent?

(a) $y = x^{e}$ . The base $x$ varies; the exponent $e \approx 2.718$ is just a constant number. This is the power rule — nothing special about the number $e$ sitting up top: $y^{'} = e x^{e - 1} .$
(b) $y = e^{x}$ . The base $e$ is constant; the exponent $x$ varies. This is an exponential, not a power. Its rule (see Derivative of exponential functions) is $\frac{d}{d x} a^{x} = a^{x} ln a$ , and since $ln e = 1$ : $y^{'} = e^{x} .$ WHY they differ: the power rule was proven by expanding $(x + h)^{n}$ in $x$ ; that machinery cannot touch a variable exponent. Different structure ⇒ different tool.

Q14. (i) Directly. For $x \neq = 0$ , $x^{0} = 1$ , a constant. The derivative of any constant is $0$ (from the limit definition, $\frac{1 - 1}{h} = 0$ for all $h$ ): $f^{'} (x) = 0.$ (ii) By the formula. $n = 0$ gives $f^{'} (x) = n x^{n - 1} = 0 \cdot x^{- 1} = 0.$ They agree. WHAT the factor of $n$ protects: even though $x^{- 1}$ is undefined at $x = 0$ , the leading factor $n = 0$ multiplies it to $0$ before that ever matters. The formula quietly encodes "a constant has zero slope" — the $n = 0$ front factor is exactly what kills the otherwise awkward $x^{- 1}$ . WHY this is the promised consistency: Stage 2 of the proof and the general formula give the same answer, so the four stages knit together seamlessly.

Recall Self-test scoreboard

Nailed L1–L2 unaided ::: You can apply the rule reliably. Nailed L3 unaided ::: You can reason about slopes and tangents, including vertical ones. Nailed L4 unaided ::: You can rebuild the rule from the limit definition and implicitly. Nailed L5 unaided ::: You spot the power-vs-exponential trap and handle degenerate cases — mastery.

Connections

Power rule — proof for integer, rational exponents — the parent theorem these drill.
Limit definition of the derivative — used in Q10, Q11, Q14(i).
Binomial theorem — expands $(x + h)^{3}$ in Q10.
Chain rule — attaches $\frac{d y}{d x}$ in Q12.
Implicit differentiation — the method of Q12.
Quotient rule — the tempting-but-wrong path warned against in Q5.
Derivative of exponential functions — the correct rule for Q13(b).