# Differentiation: Definition and Fundamental Properties

Derivatives let us pin down how fast something is changing at a single instant, not just on average over an interval. This reading builds the derivative from the ground up — starting with average rate of change, tightening that into an instantaneous rate via a limit, then developing the toolkit of rules (power, sum/difference, product, quotient, trig, exponential, logarithmic) that let you differentiate almost any combination of familiar functions without going back to the limit definition every time.

# Average and Instantaneous Rates of Change

# Average rate of change

The average rate of change of a function $f$ over an interval from $x = a$ to $x = a+h$ is a difference quotient:

$$\frac{f(a+h)-f(a)}{h}$$

Equivalently, using two endpoints $a$ and $x$:

$$\frac{f(x)-f(a)}{x-a}$$

Both expressions compute the slope of the secant line connecting the two points $(a, f(a))$ and $(x, f(x))$ on the graph of $f$. Geometrically, this is rise over run; in applied contexts, it's a rate — miles per hour, dollars per year, graduates per dollar of investment.

• as $h$ shrinks, the secant line through $(a, f(a))$ and $(a+h, f(a+h))$ rotates and starts to look like the tangent line at $a$.
• the units of a difference quotient are always (units of $f$) per (units of $x$).

# Visualizing the secant-to-tangent process

title: Secant line approaching the tangent line at x = 3, for f(x) = x^2
xlabel: x
ylabel: f(x)
bounds: 0, 6, -2, 30
y=x^2
y=6x-9
(3,9) # (3, f(3))
(4,16) # (3+h, f(3+h)), h = 1

The line $y = 6x - 9$ is the actual tangent at $x=3$; the segment through $(3,9)$ and $(4,16)$ is a secant using $h=1$. As $h \to 0$, that secant rotates onto the tangent line, and its slope approaches $6$.

# From average to instantaneous

The instantaneous rate of change at $x = a$ is what the average rate of change approaches as the interval shrinks to a single point. This is captured by a limit:

$$\lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \quad \text{or equivalently} \quad \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$$

provided the limit exists. This value is denoted $f'(a)$ — the derivative of $f$ at $a$.

example. Let $f(x) = x^2$. Estimate the instantaneous rate of change at $x = 3$ using the difference quotient with $h = 0.1$, then find the exact value.

Using $h = 0.1$:

$$\frac{f(3.1) - f(3)}{0.1} = \frac{9.61 - 9}{0.1} = 6.1$$

Exact value using the limit:

$$\lim_{h \to 0} \frac{(3+h)^2 - 9}{h} = \lim_{h \to 0} \frac{9 + 6h + h^2 - 9}{h} = \lim_{h \to 0} (6+h) = 6$$

so $f'(3) = 6$, and the estimate with $h = 0.1$ was close.

# Defining the Derivative of a Function and Using Derivative Notation

# The derivative as a function

Rather than evaluating the limit at a single point, we can define the derivative as a new function $f'$ whose value at any $x$ is:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

provided this limit exists at that $x$. This means $f'$ takes an input $x$ and outputs the slope of the tangent line to $f$ at that point — differentiation converts a function into its rate-of-change function.

flowchart TD
  A["Start with a function f(x)"] --> B["Apply the limit definition: f'(x) = lim as h to 0 of (f(x+h) - f(x)) / h"]
  B --> C{"Does the limit exist at x?"}
  C -->|"yes"| D["f is differentiable at x; f'(x) is the slope of the tangent line there"]
  C -->|"no"| E["f is not differentiable at x (corner, cusp, or vertical tangent)"]

# Notation

For $y = f(x)$, the derivative can be written several equivalent ways:

• these notations are interchangeable — problems will use whichever is convenient, and you should be comfortable switching between them.
• the derivative can also be represented graphically (the slope of a curve at each point), numerically (a table of values), and verbally (a rate described in words) — being able to move between these representations is a core AP skill.

# Tangent lines

Because $f'(a)$ is the slope of the line tangent to the graph of $f$ at $x = a$, the equation of that tangent line follows from point-slope form:

$$y - f(a) = f'(a)(x - a)$$

example. Find the tangent line to $f(x) = x^3$ at $x = 2$.

Here $f(2) = 8$. Using the definition:

$$f'(2) = \lim_{h\to 0}\frac{(2+h)^3 - 8}{h} = \lim_{h \to 0}\frac{8 + 12h + 6h^2 + h^3 - 8}{h} = \lim_{h\to 0}(12 + 6h + h^2) = 12$$

So the tangent line is $y - 8 = 12(x-2)$, or $y = 12x - 16$.

# Estimating Derivatives of a Function at a Point

Not every derivative needs to be computed analytically — many AP problems give you a table or graph and ask you to estimate $f'(a)$.

# Estimating from a table

If a table gives values of $f$ near $x = a$, approximate $f'(a)$ with a difference quotient built from the two table values closest to $a$, one on each side when possible.

{
  "title": "Cost function C(x) from table data",
  "data": {"values": [
    {"x": 3, "C": 11.2},
    {"x": 4, "C": 12.8},
    {"x": 5, "C": 13.5}
  ]},
  "mark": {"type": "line", "point": true},
  "encoding": {
    "x": {"field": "x", "type": "quantitative", "title": "x"},
    "y": {"field": "C", "type": "quantitative", "title": "C(x), in dollars"}
  }
}

example. Estimate $C'(4)$ using the symmetric difference quotient:

$$C'(4) \approx \frac{C(5)-C(3)}{5-3} = \frac{13.5 - 11.2}{2} = 1.15$$

• always show the difference-quotient structure explicitly — on the free-response section, a correct numerical answer without this structure can lose credit.
• when only one side of the table is available, use the one-sided difference quotient instead.

# Estimating from a graph

On a graph, estimate the derivative at a point by drawing (or visualizing) the tangent line there and estimating its slope from two points that the line clearly passes through.

# Using technology

Graphing calculators can numerically estimate derivatives at a point, and this is an accepted method on the exam. When you evaluate a derivative using a calculator, present the expression you evaluated and round or truncate to three places after the decimal point, storing intermediate values in the calculator to avoid compounding rounding error.

# Connecting Differentiability and Continuity

# Differentiability implies continuity

If a function is differentiable at a point, it must be continuous there. The contrapositive is often more useful in practice: if $x = a$ is not in the domain of $f$, then $a$ cannot be in the domain of $f'$ either — you cannot have a slope at a point where the function itself doesn't exist.

# Continuity does not imply differentiability

The converse is false — a function can be continuous at a point yet fail to be differentiable there. Two classic ways this happens: a corner or cusp, and a vertical tangent line.

corner example: f(x) = |x|, at x = 0

title: A corner point — f(x) = absolute value of x
xlabel: x
ylabel: f(x)
bounds: -4, 4, -1, 4
y=|x|
(0,0) # corner, not differentiable here

The slope approaching from the left is $-1$ and from the right is $+1$, so the limit defining $f'(0)$ does not exist even though $f$ is continuous at $0$.

vertical tangent example: f(x) = cube root of x, at x = 0

title: A vertical tangent — f(x) = cube root of x
xlabel: x
ylabel: f(x)
bounds: -4, 4, -2, 2
y=\operatorname{sign}(x)|x|^{1/3}
(0,0) # vertical tangent, not differentiable here

The graph is continuous at $x = 0$, but the tangent line there is vertical, so $f'(0)$ does not exist.

example. Explain why $f(x) = |x - 2|$ is not differentiable at $x = 2$.

$f$ is continuous at $x = 2$ since $f(2) = 0$ and the graph has no break there. But the difference quotient approaching from the left gives slope $-1$, while approaching from the right gives slope $+1$. Since these one-sided limits disagree, $\displaystyle\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}$ does not exist, so $f$ is not differentiable at $x = 2$ — this is a corner point.

# Applying the Power Rule

For functions of the form $f(x) = x^r$, where $r$ is any real number, the power rule states:

$$\frac{d}{dx}\left[x^r\right] = rx^{r-1}$$

This can be derived directly from the limit definition of the derivative for integer powers, and it holds more generally for any real exponent, including negative and fractional exponents.

title: f(x) = x squared and its derivative f'(x) = 2x
xlabel: x
ylabel: y
bounds: -4, 4, -10, 10
f(x)=x^2
g(x)=2x

Notice how $f'(x) = 2x$ is negative wherever $f$ is decreasing, zero at the minimum of $f$, and positive wherever $f$ is increasing — the sign of the derivative tracks the direction of the original function.

example. Differentiate $g(x) = x^{-3}$ and $h(x) = \sqrt{x}$.

$$g'(x) = -3x^{-4}$$

Rewriting $h(x) = x^{1/2}$:

$$h'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$$

# Derivative Rules: Constant, Sum, Difference, and Constant Multiple

Beyond the power rule, four structural rules let you differentiate combinations of functions term by term.

• the derivative of a constant is zero: $\frac{d}{dx}[c] = 0$.
• constant multiple rule: $\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$.
• sum rule: $\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$.
• difference rule: $\frac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)$.

Combined with the power rule, these are enough to differentiate any polynomial term by term.

example. Differentiate $p(x) = 4x^5 - 7x^2 + 9x - 3$.

$$p'(x) = 20x^4 - 14x + 9$$

Note the constant term $-3$ contributes nothing, since its derivative is $0$.

# Derivatives of cos x, sin x, e^x, and ln x

Certain families of functions have derivatives that must simply be known — they don't reduce to the power rule.

$$\frac{d}{dx}[\sin x] = \cos x \qquad \frac{d}{dx}[\cos x] = -\sin x$$

$$\frac{d}{dx}[e^x] = e^x \qquad \frac{d}{dx}[\ln x] = \frac{1}{x}$$

The function $e^x$ is unique in being its own derivative, which is why it appears so often in models of growth and decay.

title: y = e^x compared with y = ln(x)
xlabel: x
ylabel: y
bounds: -4, 6, -4, 8
f(x)=e^x
g(x)=\ln(x)
y=x

The two curves are mirror images across the line $y=x$, since $\ln x$ and $e^x$ are inverse functions of each other.

# Using the derivative definition to evaluate limits

Sometimes a limit that looks difficult is secretly the definition of a derivative in disguise. Recognizing this structure avoids messy limit algebra entirely.

example. Evaluate $\displaystyle\lim_{h \to 0} \frac{\sin\left(\frac{\pi}{6}+h\right) - \sin\left(\frac{\pi}{6}\right)}{h}$.

This has exactly the form $\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ with $f(x) = \sin x$ and $a = \frac{\pi}{6}$. So the limit equals $f'\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.

# The Product Rule

For two differentiable functions $u(x)$ and $v(x)$, the derivative of their product is not simply the product of their derivatives. Instead:

$$\frac{d}{dx}\left[u(x)v(x)\right] = u'(x)v(x) + u(x)v'(x)$$

• a common error is differentiating each factor separately and multiplying the results together — this does not work and should be actively avoided.
• when evaluating a product rule derivative at a specific point, show the full structure with numbers substituted in before simplifying, e.g. $f'(3) = u'(3)v(3) + u(3)v'(3)$.

example. Let $f(x) = x^2 \sin x$. Find $f'(x)$.

With $u(x) = x^2$, $u'(x) = 2x$, $v(x) = \sin x$, $v'(x) = \cos x$:

$$f'(x) = 2x\sin x + x^2\cos x$$

# The Quotient Rule

For a quotient of differentiable functions where $v(x) \neq 0$:

$$\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$$

• the order of terms in the numerator matters, since subtraction is not commutative — a common memory aid is "low d-high minus high d-low, over low squared."
• as with the product rule, show the full substituted structure at a point before simplifying.

example. Let $f(x) = \frac{x^2}{\cos x}$. Find $f'(x)$.

With $u(x) = x^2$, $u'(x) = 2x$, $v(x) = \cos x$, $v'(x) = -\sin x$:

$$f'(x) = \frac{2x\cos x - x^2(-\sin x)}{\cos^2 x} = \frac{2x\cos x + x^2 \sin x}{\cos^2 x}$$

# Finding the Derivatives of Tangent, Cotangent, Secant, and Cosecant Functions

Because $\tan x$, $\cot x$, $\sec x$, and $\csc x$ can each be rewritten as a ratio involving $\sin x$ and $\cos x$, their derivatives follow directly from the quotient rule applied to those identities.

$$\frac{d}{dx}[\tan x] = \sec^2 x \qquad \frac{d}{dx}[\cot x] = -\csc^2 x$$

$$\frac{d}{dx}[\sec x] = \sec x \tan x \qquad \frac{d}{dx}[\csc x] = -\csc x \cot x$$

title: y = tan(x) and its derivative y = sec^2(x)
xlabel: x (radians)
ylabel: y
bounds: -3, 3, -6, 6
f(x)=\tan(x)
g(x)=\sec(x)^2

example. Derive $\frac{d}{dx}[\tan x]$ from the quotient rule.

Write $\tan x = \frac{\sin x}{\cos x}$. With $u = \sin x$, $u' = \cos x$, $v = \cos x$, $v' = -\sin x$:

$$\frac{d}{dx}[\tan x] = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$$

# Exercises

1. Let $f(x) = x^2 - 3x$. Use the limit definition of the derivative to find $f'(x)$.
2. A table gives $g(2) = 5.1$, $g(3) = 6.4$, $g(4) = 8.0$. Estimate $g'(3)$ using a symmetric difference quotient.
3. Explain why $f(x) = \sqrt[3]{x-1}$ is not differentiable at $x = 1$.
4. Differentiate $h(x) = 5x^4 - 2x^{-1} + \sqrt{x}$.
5. Find $\frac{d}{dx}\left[3\cos x - 2e^x + \ln x\right]$.
6. Let $f(x) = (3x^2 + 1)(\tan x)$. Find $f'(x)$.
7. Let $f(x) = \frac{e^x}{x^2 + 1}$. Find $f'(x)$.
8. Find an equation for the line tangent to $f(x) = x^3 - 2x$ at $x = 1$.
9. Differentiate $y = \sec x - \cot x$.
10. Given $f(x) = u(x)v(x)$ with $u(2) = 4$, $u'(2) = -1$, $v(2) = 3$, $v'(2) = 2$, find $f'(2)$.