Visual walkthrough — Power rule — proof for integer, rational exponents

Step 0 — What is a derivative, as a picture?

PICTURE: Two points sit on the curve. The horizontal gap between them is $h$. The vertical gap is $f(x+h)-f(x)$. The straight line joining them (the secant) has slope $\underset{\text{rise}÷\text{run}}{\underbrace{\frac{f(x+h)-f(x)}{h}}}.$ As we slide the second point toward the first ($h\to 0$), the secant tips over into the tangent — the line that just grazes the curve.

Every step below feeds a specific $f$ into this one machine. See Limit definition of the derivative.

Step 1 — The area picture for $x^2$ (why "bring the two down")

WHY a square? Because "$x$ squared" is a square, and growing a square is something we can literally draw. When the side grows from $x$ to $x+h$, the new area is $(x+h{)}^2$. The extra area is a thin L-shaped border.

PICTURE: The border splits into three pieces:

• a strip along the top: width $x$, height $h$ → area $xh$,
• a strip along the side: width $h$, height $x$ → area $xh$,
• a tiny corner square: $h\times h$ → area $h^2$.

$\underset{\text{extra area}}{\underbrace{(x+h{)}^2-x^2}}=\underset{\text{two long strips}}{\underbrace{xh+xh}}+\underset{\text{tiny corner}}{\underbrace{h^2}}=2xh+h^2.$

Divide by the run $h$: $\frac{2xh+h^2}{h}=2x+h.$ Now shrink $h\to 0$: the tiny corner term $h$ vanishes, leaving $f^{\prime }(x)=2x$. The "$2$" came from the two long strips — there are two of them because a square has two dimensions. This is the power rule $n{x}^{n-1}$ with $n=2$: seen, not memorized.

Step 2 — The general shape $(x+h{)}^n$ (why the binomial theorem)

WHY the binomial theorem? It is exactly the tool that expands $(x+h{)}^n$ into a sorted list of terms, organised by how many factors of $h$ each carries. That sorting is the whole point: we want the pieces with exactly one $h$, because those are the ones that survive dividing by $h$ and then $h\to 0$.

PICTURE: For $n=3$ (a cube of side $x$), the extra volume when the side grows by $h$ is 3 flat faces (each $x^2\cdot h$ → total $3x^{2}h$), plus 3 thin edges ($x{h}^2$), plus 1 corner ($h^3$). The three faces are the accent — they are the $n{x}^{n-1}h$ term with $n=3$.

Step 3 — Run it through the machine (positive integer $n$)

WHY it collapses: the $x^n$ at the front cancels the $-x^n$ from $-f(x)$; after dividing by $h$, only one term has no leftover $h$.

$\frac{(x+h{)}^n-x^n}{h}=\frac{\cancel{x^n}+n{x}^{n-1}h+\left(\genfrac{}{}{0ex}{}{n}{2}\right)x^{n-2}{h}^2+\cdots -\cancel{x^n}}{h}=n{x}^{n-1}+\underset{\text{every term still has an }h}{\underbrace{\left(\genfrac{}{}{0ex}{}{n}{2}\right)x^{n-2}h+\cdots }}.$

PICTURE: a number line for the terms after dividing by $h$. The single term $n{x}^{n-1}$ sits alone with no $h$ (accent). All other terms carry at least one $h$ and slide to $0$ as the arrow pushes $h\to 0$.

$\boxed{f^{\prime }(x)=n{x}^{n-1}}(n=1,2,3,\dots )$

Step 4 — The flat edge case $n=0$

PICTURE: the graph of $y=1$: a level line. Nudge $x$ by $h$; the height does not change, so rise $=0$, so slope $=0$. The formula agrees: $0\cdot x^{-1}=0$. ✓

This is not a special exception — it is the same rule reading "there are $0$ faces to grow."

Step 5 — Negative exponent $n=-m$ (the fraction picture)

WHY a new move? The binomial theorem only expands positive whole powers. So we cannot expand $x^{-m}$. Instead we write it as a fraction and combine over a common denominator, which re-exposes the positive-power difference $x^m-(x+h{)}^m$ we already know how to handle.

$\frac{1}{h}\left(\frac{1}{(x+h{)}^m}-\frac{1}{x^m}\right)=\frac{\stackrel{\text{Step 3 handles this}}{\overbrace{x^m-(x+h{)}^m}}}{h\underset{\to x^{2m}}{\underbrace{(x+h{)}^m{x}^m}}}.$

• top: same difference as Step 3 but flipped in sign, so it tends to $-m{x}^{m-1}$.
• bottom: as $h\to 0$, $(x+h{)}^m{x}^m\to x^{2m}$.

$f^{\prime }(x)=\frac{-m{x}^{m-1}}{x^{2m}}=-m{x}^{-m-1}=n{x}^{n-1}(n=-m).$

PICTURE: the curve $y=1/x$ ($m=1$) with a tangent line drawn at a point — its slope is a downward (negative) ramp, matching $-1\cdot x^{-2}<0$.

Step 6 — Rational exponent $n=p/q$ (the flip-game)

WHY implicit differentiation? If we name the height $y=x^{p/q}$ and raise both sides to the power $q$, the fraction dies: $y^q=x^p$. Now both sides are integer powers — territory we already conquered in Steps 3–5. We differentiate both sides, using the Chain rule on the left because $y$ is itself a function of $x$. See Implicit differentiation.

$\frac{d}{dx}{y}^q=\frac{d}{dx}{x}^p⟹\underset{\text{chain rule: }y\text{ depends on }x}{\underbrace{q{y}^{q-1}\frac{dy}{dx}}}=\underset{\text{plain power rule}}{\underbrace{p{x}^{p-1}}}.$

Solve for the slope $\frac{dy}{dx}$, then substitute $y=x^{p/q}$ (so $y^{q-1}=x^{p-p/q}$):

$\frac{dy}{dx}=\frac{p}{q}\cdot \frac{x^{p-1}}{x^{p-p/q}}=\frac{p}{q}{x}^{p/q-1}=n{x}^{n-1}.$

PICTURE: the graph of $y=\sqrt{x}$. It rises steeply near $0$ and flattens as $x$ grows — its tangent slopes match $\frac{1}{2\sqrt{x}}$, which is huge near $0$ and small far out.

The one-picture summary

Every stage feeds the same limit machine; only the algebra to simplify the ratio changes, and each later stage reuses an earlier one.

Active recall