Defining the Derivative of a Function and Using Derivative Notation

AP Calculus · Differentiation: Definition and Fundamental Properties

Defining the Derivative and Its Notation

A secant slope with the gap shrunk to nothing — written five different ways.

key idea$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

What it says

The derivative at a point

\[f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\]

The difference quotient $\dfrac{f(a+h)-f(a)}{h}$ is the slope of the secant through $(a,f(a))$ and $(a+h,f(a+h))$. Letting $h\to0$ turns that secant into the tangent, and its slope is $f'(a)$.

The derivative as a function

\[f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]

Leave $a$ as a variable and the same limit produces a new function whose output at each $x$ is the slope there. This is why one computation serves every point at once.

secant over $[1,3]$

secant over $[1,1.6]$

tangent at $x=1$

The same curve three times. As the second point slides toward $x=1$, the secant pivots and settles on the tangent. The derivative is the number that pivoting converges to.

Reading the notation

$f'(x)$Lagrange notation. Compact; the standard on multiple choice.

$\dfrac{dy}{dx}$Leibniz notation. Carries the variables, so it is preferred in applied and related-rates work where units matter.

$\dfrac{d}{dx}\left[f(x)\right]$The operator form — ‘take the derivative of what follows’. Use it when you are about to differentiate, not for the result.

$\left.\dfrac{dy}{dx}\right|_{x=a}$Leibniz evaluated at a point — the equivalent of $f'(a)$.

$y'$Same as $f'(x)$ when the function is written $y=\dots$

The alternate form$f'(a)=\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}$ — identical content, with $x$ sliding toward $a$ instead of $h$ shrinking to $0$.

what the symbols mean

$h$ is a gap, not a point — it shrinks to zero but never equals it.
The quotient is always $\tfrac00$ at $h=0$; the limit is what rescues it.
Units of $f'$ = units of output per unit of input.
$f'(a)$ is a number; $f'(x)$ is a function.

Worked examples

EX 1

Derivative at a point from the definition

$f(x)=x^{2}$ at $a=3$

Use the limit definition to find $f'(3)$ for $f(x)=x^{2}$.

Write the difference quotient

$\dfrac{f(3+h)-f(3)}{h}=\dfrac{(3+h)^{2}-9}{h}$.

Expand

$(3+h)^{2}=9+6h+h^{2}$, so the numerator is $6h+h^{2}$.

Cancel the $h$

$\dfrac{6h+h^{2}}{h}=6+h$ — legal because $h\neq0$ throughout a limit.

Take the limit

$\lim_{h\to0}(6+h)=6$.

State it

$f'(3)=6$: the tangent to $y=x^{2}$ at $x=3$ has slope $6$.

EX 2

Derivative as a function

$f(x)=x^{2}$

Now find $f'(x)$ for general $x$.

Same set-up, variable point

$\dfrac{(x+h)^{2}-x^{2}}{h}$.

Expand and simplify

$\dfrac{x^{2}+2xh+h^{2}-x^{2}}{h}=\dfrac{2xh+h^{2}}{h}=2x+h$.

Take the limit

$\lim_{h\to0}(2x+h)=2x$.

Compare with Example 1

Substituting $x=3$ gives $6$ — consistent, and now every point is covered at once.

Note what changed

Only whether the point was fixed. One extra symbol converts a number into a function.

EX 3

Recognise a limit as a derivative

$\lim\limits_{h\to0}\dfrac{\sqrt{4+h}-2}{h}$

Evaluate $\lim\limits_{h\to0}\dfrac{\sqrt{4+h}-2}{h}$ without doing any algebra.

Match the pattern

It has the shape $\dfrac{f(a+h)-f(a)}{h}$ with $f(x)=\sqrt{x}$ and $a=4$, since $f(4)=2$.

Name what the limit is

By definition, this limit is $f'(4)$.

Use the known derivative

$f'(x)=\dfrac{1}{2\sqrt x}$, so $f'(4)=\dfrac{1}{2\cdot2}=\dfrac14$.

State the answer

$\dfrac14$.

Why this is worth spotting

These appear on multiple choice disguised as hard limits. Recognising the form turns a page of algebra into one line.

EX 4

Units and interpretation

applied

$C(q)$ is the cost in dollars of producing $q$ widgets, and $C'(200)=13$. Interpret this.

Read the units

$C$ is in dollars and $q$ in widgets, so $C'$ is in dollars per widget.

Say what the number means

At a production level of $200$ widgets, cost is increasing at about $13$ dollars per additional widget.

Give the practical reading

The $201$st widget costs roughly $13 to make.

Note the approximation

‘Roughly’ matters: $C'(200)$ is an instantaneous rate, while the actual next widget spans a whole unit.

The habit

Always state a derivative in context as output units per input unit, at the stated input.

Boundary cases

$h\to0$ is not $h=0$

Setting $h=0$ gives $\tfrac00$. The limit exists precisely because $h$ approaches zero without arriving — which is what licenses the cancellation.

$f'(a)$ and $f'(x)$ are different objects

One is a number, the other a function. Writing $f'(3)=2x$ mixes them; the correct statement is $f'(x)=2x$, so $f'(3)=6$.

The alternate form is the same definition

$\lim_{x\to a}\tfrac{f(x)-f(a)}{x-a}$ measures the same slope with $x$ sliding in. Use whichever the question hands you.

Notation carries information

$\tfrac{dy}{dx}$ names both variables, which is why it is the right choice in applied contexts. $f'$ is silent about units.

Course map. Topic 2.2 · LO CHA-2.B — represent the derivative of a function as a limit of difference quotients (EK CHA-2.B.1, CHA-2.B.2, CHA-2.B.3, CHA-2.B.4).