Differentiation: Definition and Fundamental Properties

What a Derivative Measures

A derivative is calculus’ way of describing how something changes right now. In algebra, you describe patterns with formulas. In calculus, you go one level deeper and ask how sensitive the output is to small changes in the input.

There are two big, equivalent interpretations you should build from the start:

1. Slope interpretation (geometric): the derivative is the slope of the tangent line to a graph at a point.
2. Rate interpretation (applied): the derivative is the instantaneous rate of change of one quantity with respect to another.

AP problems move back and forth between these interpretations constantly.

Rates of change: average vs instantaneous

There are two main ways to talk about “rate of change.”

Over an interval from $x=a$ to $x=b$, the average rate of change is

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

In coordinate form, this is the familiar slope formula

$\frac{y_2-y_1}{x_2-x_1}$

In many contexts, this is “rate of change over an interval of time.”

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Geometrically, $\frac{f(b)-f(a)}{b-a}$ is the slope of the **secant line** through the two points $\left(a,f(a)\right)$ and $\left(b,f(b)\right)$.

To capture change at a specific point in time, you use the instantaneous rate of change, which comes from the same difference quotient but with a limit as $h \to 0$.

Slopes on curves: secant lines and tangent lines

For a linear function, slope is “rise over run” and is constant. For a curved graph, the slope changes from point to point, so you approximate the slope near a point using a secant line.

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The closer the two points are, the more accurate the secant slope becomes as an estimate of the “true” slope at that point.

A tangent line is the limiting position of these secant lines as the second point moves in toward the first. It is commonly described as a line that “touches the curve at exactly one point,” but the most reliable definition is this: it is the line whose slope equals the limiting slope of nearby secants (and a function can cross its tangent line).

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From average change to instantaneous change (difference quotient)

Pick a nearby point $a+h$ and compute the slope from $a$ to $a+h$:

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

As $h$ gets closer to $0$, the second point slides toward the first point. If the slopes approach a single value, that limiting slope is the derivative.

Real-world meaning (why you should care)

Derivatives show up wherever a small input change produces an output change:

• Motion: if $s(t)$ is position, then $s'(t)$ is instantaneous velocity.
• Economics: if $C(x)$ is cost to produce $x$ items, then $C'(x)$ is marginal cost (approximate cost of producing one more item).
• Population: if $P(t)$ is population, then $P'(t)$ is the instantaneous growth rate.

The key idea is that a derivative is a local measurement. It tells you what the function is doing near a point.

Example: average rate of change as a secant slope

Let $f(x)=x^2$. Compute the average rate of change from $x=1$ to $x=3$.

$\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4$

That means over that interval, $f(x)$ increases about 4 units in output per 1 unit in input.

Example: zooming in toward an instantaneous rate

Using the same $f(x)=x^2$, look at slopes from $x=2$ to $x=2+h$:

$\frac{f(2+h)-f(2)}{h}=\frac{(2+h)^2-4}{h}=\frac{4+4h+h^2-4}{h}=\frac{4h+h^2}{h}=4+h$

If $h$ is very small, the slope is very close to $4$. The limiting value as $h \to 0$ will be the derivative at $x=2$.

Exam Focus

• Typical question patterns
  • Given a table or context, compute an average rate of change and interpret units.
  • Given a graph, estimate slopes of secant lines and reason what happens as points get closer.
  • Translate wording like “instantaneous rate at $x=a$” into the limit idea.
• Common mistakes
  • Mixing up $\frac{f(b)-f(a)}{b-a}$ (average on an interval) with the derivative (instantaneous at a point).
  • Forgetting that the denominator in $\frac{f(a+h)-f(a)}{h}$ is the change in input (so it must match the function’s input variable and units).
  • Treating “tangent line” as a line that merely touches once; many functions cross their tangent line.

The Limit Definition of the Derivative

The derivative is defined using a limit because we need a precise way to capture what happens as the interval shrinks to zero. The limit ensures we’re not just picking a tiny interval; we’re describing what value the slopes approach.

Definition at a point

The derivative of $f$ at $x=a$, written $f'(a)$, is

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

This expression is called a difference quotient. It is the slope of a secant line, taken to the limit where the secant becomes a tangent.

There’s an equivalent form you’ll also see:

$f'(a)=\lim_{x \to a}\frac{f(x)-f(a)}{x-a}$

These represent the same idea with different approach variables.

Derivative as a function

If you replace the specific number $a$ with a variable $x$, you get the derivative function:

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

This matters because problems often ask for an expression for $f'(x)$ (a new function), not just a single slope at one point.

When the derivative does not exist

The limit must exist as a real number. If the left-hand and right-hand approaches disagree, or if the slope becomes infinite, then $f'(a)$ does not exist.

Common reasons a derivative fails:

• discontinuity (a jump, hole, or asymptote)
• corner or cusp (left and right slopes don’t match or blow up differently)
• vertical tangent (slope approaches $\pm\infty$)

Worked example: derivative from the definition

Find $f'(x)$ for $f(x)=x^2$ using the limit definition.

$f'(x)=\lim_{h \to 0}\frac{(x+h)^2-x^2}{h}$

$f'(x)=\lim_{h \to 0}\frac{x^2+2xh+h^2-x^2}{h}=\lim_{h \to 0}\frac{2xh+h^2}{h}$

$f'(x)=\lim_{h \to 0}(2x+h)=2x$

Worked example: derivative at a point from the definition (conjugates)

Let $f(x)=\sqrt{x}$. Find $f'(4)$ from first principles.

$f'(4)=\lim_{h \to 0}\frac{\sqrt{4+h}-\sqrt{4}}{h}=\lim_{h \to 0}\frac{\sqrt{4+h}-2}{h}$

Multiply by the conjugate:

$f'(4)=\lim_{h \to 0}\frac{\sqrt{4+h}-2}{h}\cdot\frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}$

$f'(4)=\lim_{h \to 0}\frac{(4+h)-4}{h(\sqrt{4+h}+2)}=\lim_{h \to 0}\frac{h}{h(\sqrt{4+h}+2)}$

$f'(4)=\lim_{h \to 0}\frac{1}{\sqrt{4+h}+2}=\frac{1}{4}$

Interpreting: near $x=4$, $\sqrt{x}$ increases at about 0.25 output units per 1 input unit.

Exam Focus

• Typical question patterns
  • “Use the definition of the derivative to find $f'(x)$” for a simple function.
  • Compute $f'(a)$ exactly from the limit definition (often with algebraic simplification like conjugates).
  • Decide whether a derivative exists at a point based on one-sided limits.
• Common mistakes
  • Plugging in $h=0$ too early (you must simplify before evaluating the limit).
  • Losing parentheses when expanding $f(x+h)$ (a very common source of sign errors).
  • Forgetting to use the conjugate when radicals cause an indeterminate form.

Derivative Notation and the Meaning of $\frac{dy}{dx}$

Calculus uses multiple notations for derivatives because they emphasize different interpretations. You should be able to read all of them fluently.

Notation you will see (and what it means)

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

You will also see second derivative notation, which measures how the first derivative changes:

What $\frac{dy}{dx}$ is (conceptually)

The notation $\frac{dy}{dx}$ comes from the idea “change in output over change in input,” because derivatives start from the difference quotient and then become instantaneous via a limit. In a strict algebraic sense, $\frac{dy}{dx}$ is not simply a fraction you can always treat like ordinary division, but it often behaves like one in later topics.

For Unit 2, the safest interpretation is:

• $\frac{dy}{dx}$ represents the slope of the tangent line to the graph of $y$ versus $x$.
• Units of $\frac{dy}{dx}$ are “units of $y$ per unit of $x$.”

Derivative values vs derivative function

It’s important not to blur these two:

• $f'(a)$ is a single number (slope at $x=a$).
• $f'(x)$ is a new function giving the slope at every $x$ where it exists.

Example: interpreting units

If $s(t)$ measures distance in meters and $t$ is time in seconds, then $s'(t)$ has units meters per second.

If $C(x)$ measures cost in dollars and $x$ is number of items, then $C'(x)$ has units dollars per item.

Example: tangent slope from notation

Suppose you’re told $f'(3)=-2$. That means the tangent line slope at $x=3$ is $-2$, and near $x=3$, increasing $x$ by about 1 decreases $f(x)$ by about 2.

Exam Focus

• Typical question patterns
  • Given $f'(a)$, interpret what it says about the function’s behavior and include units.
  • Convert between notations: recognize that $\frac{d}{dx}[f(x)]$ means “differentiate $f(x)$.”
  • Use derivative information to write a tangent line equation (often paired with a function value).
• Common mistakes
  • Treating $f'(x)$ and $f(x)$ as interchangeable (they are different functions).
  • Dropping units in context problems (units are part of the meaning of a derivative).
  • Confusing $f'(a)$ with $f(a)$, especially when both are provided in a word problem.

Differentiability and Continuity

The derivative is defined by a limit, so differentiability is closely linked to continuity, but they are not the same idea.

What it means to be differentiable

A function $f$ is differentiable at $x=a$ if $f'(a)$ exists as a finite real number. It is differentiable on an interval if it’s differentiable at every point in that interval.

Geometrically, differentiable at a point means the graph has a well-defined (finite) tangent slope there.

Differentiability implies continuity

A fundamental fact is:

• If $f$ is differentiable at $x=a$, then $f$ is continuous at $x=a$.

But the converse is not guaranteed:

• A function can be continuous at $a$ but not differentiable at $a$.

Where differentiability fails (and what it looks like)

Even if a function is continuous, the slope might fail to settle to one value.

1. Corner: left-hand slope and right-hand slope are finite but unequal (for example, $|x|$ at $0$).
2. Cusp: the slope becomes infinite in different ways from left and right.
3. Vertical tangent: slopes approach $\pm\infty$ from both sides in the same way.
4. Discontinuity: if the function is not continuous, it cannot be differentiable.

A precise way to test “left vs right” uses one-sided derivatives:

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

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

If both exist and are equal, then $f'(a)$ exists.

Example: a continuous function that is not differentiable

Let $f(x)=|x|$. Compute slopes from the definition at $x=0$.

For $h>0$:

$\frac{f(0+h)-f(0)}{h}=\frac{|h|-0}{h}=\frac{h}{h}=1$

For $h<0$:

$\frac{f(0+h)-f(0)}{h}=\frac{|h|-0}{h}=\frac{-h}{h}=-1$

The right-hand derivative is 1 and the left-hand derivative is -1, so the derivative does not exist at 0.

Example: vertical tangent (derivative does not exist as a finite number)

Consider $f(x)=x^{1/3}$ at $x=0$:

$f'(0)=\lim_{h \to 0}\frac{h^{1/3}-0}{h}=\lim_{h \to 0}h^{-2/3}$

As $h \to 0$, $h^{-2/3}$ grows without bound, so the slope becomes infinite, indicating a vertical tangent line at the origin.

Exam Focus

• Typical question patterns
  • Given a graph, identify where $f'(x)$ does not exist and justify using corners, cusps, vertical tangents, or discontinuities.
  • Use continuity and differentiability logic: “differentiable implies continuous” to eliminate answer choices.
  • Evaluate one-sided derivative limits to test differentiability.
• Common mistakes
  • Assuming continuity automatically guarantees differentiability.
  • Calling a vertical tangent “differentiable with a very large slope” (the derivative must be a finite real number).
  • Forgetting that sharp points can occur even in piecewise formulas that look algebraic.

Linearity Properties and the Power Rule

Computing derivatives from the limit definition is conceptually important, but it’s too slow for most functions you care about. Derivative rules are shortcuts justified by the limit definition.

Linearity: constants, sums, and constant multiples

If $c$ is a constant and $f$ and $g$ are differentiable, then:

Constant rule

$\frac{d}{dx}[c]=0$

Example: if $f(x)=10$, then $f'(x)=0$.

Constant multiple rule

$\frac{d}{dx}[cf(x)]=cf'(x)$

Sum and difference rules

$\frac{d}{dx}[f(x)+g(x)]=f'(x)+g'(x)$

$\frac{d}{dx}[f(x)-g(x)]=f'(x)-g'(x)$

The power rule

For $n$ a positive integer,

$\frac{d}{dx}[x^n]=nx^{n-1}$

A common way to remember it is: multiply down and decrease the power.

Examples:

• $x^4$ becomes $4x^3$
• $2x^2$ becomes $4x$

The power rule works especially well for polynomials, and it also extends to negative integer powers (and more general exponents later).

Differentiating polynomials

A polynomial is a sum of constant multiples of powers of $x$, so you differentiate term-by-term.

Example: derivative of a polynomial

Differentiate

$f(x)=3x^4-5x^2+7x-9$

Then

$f'(x)=12x^3-10x+7$

Example: using a derivative to find a slope

Let $g(x)=x^3-2x$. Find the slope of the tangent at $x=-1$.

$g'(x)=3x^2-2$

$g'(-1)=3(1)-2=1$

A common misconception: “power rule for everything”

It’s tempting to treat expressions like $\sqrt{x+1}$ as if the inside doesn’t matter and apply the power rule directly. In Unit 2, you should mainly apply these rules to sums of simple powers of $x$. More complicated inputs are handled systematically later (for example, with the chain rule).

Exam Focus

• Typical question patterns
  • Differentiate polynomials quickly and evaluate at a point.
  • Given $f'(a)$ or a derivative expression, interpret or use it in a tangent-line setup.
  • Identify which derivative rules justify each differentiation step.
• Common mistakes
  • Forgetting to reduce the exponent by 1 in the power rule.
  • Losing negative signs when differentiating term-by-term.
  • Treating constants like 7 as if they differentiate to 7 instead of 0.

Product and Quotient Rules

Linearity and the power rule work for sums of terms, but many functions are built by multiplying or dividing expressions. Derivatives do not distribute over multiplication or division the way they distribute over addition.

Product rule

If $h(x)=f(x)g(x)$, then

$h'(x)=f'(x)g(x)+f(x)g'(x)$

A common memory aid is 1d2 + 2d1 (first times derivative of second, plus second times derivative of first).

This also highlights why the incorrect shortcut below is not true:

$\frac{d}{dx}[f(x)g(x)]=f'(x)g'(x)$

If you have two polynomials multiplied by each other like $((2x+7)(9x+8))$, you could multiply it out and then use the power rule, but the product rule is usually faster and less error-prone.

Example: product rule in action

Differentiate

$h(x)=(x^2+1)(3x-4)$

Let $f(x)=x^2+1$ and $g(x)=3x-4$. Then

$f'(x)=2x$

$g'(x)=3$

So

$h'(x)=2x(3x-4)+(x^2+1)3$

Simplifying gives

$h'(x)=9x^2-8x+3$

Quotient rule

If

$h(x)=\frac{f(x)}{g(x)}$

then

$h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$

A common memory aid is low d high minus high d low, over low squared.

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Operationally, the key is careful substitution and parentheses to control signs.

Example: quotient rule in action

Differentiate

$q(x)=\frac{x^2+1}{x-2}$

Let $f(x)=x^2+1$ and $g(x)=x-2$. Then

$f'(x)=2x$

$g'(x)=1$

So

$q'(x)=\frac{2x(x-2)-(x^2+1)(1)}{(x-2)^2}$

Simplifying the numerator gives

$q'(x)=\frac{x^2-4x-1}{(x-2)^2}$

Domain and differentiability reminders with quotients

If $g(x)=0$ at some point, then $\frac{f(x)}{g(x)}$ is not defined there, so it cannot be differentiable there. Problems sometimes hide this: you may compute a derivative expression, but you still need to remember the original function’s domain.

Exam Focus

• Typical question patterns
  • Differentiate products of polynomials (often easiest if you use product rule rather than expand, but either can work).
  • Differentiate rational functions and evaluate the derivative at a point.
  • Determine where a derivative is undefined due to zeros in a denominator.
• Common mistakes
  • Using the incorrect shortcut that the derivative of a product is the product of the derivatives.
  • In the quotient rule, swapping the subtraction order (the numerator is $f'g-fg'$).
  • Forgetting parentheses when substituting expressions for $f$ and $g$, causing sign errors.

Derivatives of Trigonometric and Other “Memory” Functions

Trigonometric derivatives are core in calculus because they connect geometry, periodic motion, and limits. On top of that, there are a few derivatives that are usually easier to memorize than to re-derive during an exam.

A crucial detail: standard trig derivative formulas assume angles are measured in radians, not degrees.

Key trig derivative facts (radians)

$\frac{d}{dx}[\sin(x)]=\cos(x)$

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

You can then build additional derivatives using product and quotient rules, such as

$\frac{d}{dx}[\tan(x)]=\sec^2(x)$

(when $\tan(x)$ is defined).

Why radians matter (conceptual)

The derivative of $\sin(x)$ is tied to the fundamental limit

$\lim_{h \to 0}\frac{\sin(h)}{h}=1$

This limit is true when $h$ is in radians.

Example: differentiating a trig expression

Differentiate

$f(x)=3\sin(x)-2\cos(x)$

Then

$f'(x)=3\cos(x)+2\sin(x)$

Example: tangent slope on a trig graph

Let $g(x)=\cos(x)$. Find the slope of the tangent line at

$x=\frac{\pi}{3}$

$g'(x)=-\sin(x)$

$g'\left(\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}$

Example: deriving the derivative of $\tan(x)$ using the quotient rule

Since

$\tan(x)=\frac{\sin(x)}{\cos(x)}$

use the quotient rule with $f(x)=\sin(x)$ and $g(x)=\cos(x)$. Then

$f'(x)=\cos(x)$

$g'(x)=-\sin(x)$

So

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

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

Using $\sin^2(x)+\cos^2(x)=1$,

$\frac{d}{dx}[\tan(x)]=\sec^2(x)$

Other common “memory derivatives”

Some derivatives are commonly memorized because they appear so often:

• derivatives of $\sin(x)$ and $\cos(x)$ (above)
• derivative of $e^x$
• derivative of $\ln(x)$

In particular,

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

and

$\frac{d}{dx}[\ln(x)]=\frac{1}{x}$

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Exam Focus

• Typical question patterns
  • Differentiate expressions combining polynomials and $\sin(x)$ or $\cos(x)$.
  • Evaluate derivatives at special angles like $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$, $\frac{\pi}{2}$.
  • Use trig derivatives inside product and quotient rules.
• Common mistakes
  • Using degrees instead of radians when interpreting values near zero or using special angles.
  • Forgetting the negative sign in the cosine derivative.
  • Differentiating $\tan(x)$ as $\sec(x)$ (confusing function relationships with derivative relationships).

Tangent Lines, Normal Lines, and Local Linear Approximation

Once you can compute derivatives, you can use them to build linear models that approximate functions near a point. This connects algebra (lines) to calculus (derivatives) and is heavily tested.

Equation of the tangent line

At $x=a$, the slope of the tangent line is $f'(a)$, and the point on the curve is $\left(a,f(a)\right)$. Using point-slope form:

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

Equation of the normal line

The normal line is perpendicular to the tangent line. If the tangent slope is $m=f'(a)$ and $m \neq 0$, then the normal slope is

$m_{\text{normal}}=-\frac{1}{m}$

and the normal line is

$y-f(a)=m_{\text{normal}}(x-a)$

If the tangent line is horizontal (slope 0), then the normal line is vertical and cannot be written in slope-intercept form.

Local linearity and linearization

Near $x=a$, a differentiable function behaves almost like a line. The tangent line approximation is called the linearization:

$L(x)=f(a)+f'(a)(x-a)$

For $x$ close to $a$,

$f(x) \approx L(x)$

Worked example: tangent line

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

$f'(x)=2x$

$f'(3)=6$

$f(3)=9$

So the tangent line is

$y-9=6(x-3)$

which simplifies to

$y=6x-9$

Worked example: linear approximation

Approximate $\sqrt{4.1}$ using linearization.

Let $f(x)=\sqrt{x}$ and choose $a=4$.

$f(4)=2$

$f'(4)=\frac{1}{4}$

So

$L(x)=2+\frac{1}{4}(x-4)$

Then

$L(4.1)=2+\frac{1}{4}(0.1)=2.025$

So

$\sqrt{4.1} \approx 2.025$

A conceptual pitfall: “tangent line equals function”

A tangent line is not the function; it’s the best linear approximation near the point of tangency. Farther away, the curve can bend away significantly.

Exam Focus

• Typical question patterns
  • Given $f(a)$ and $f'(a)$, write the tangent line equation at $x=a$.
  • Use the tangent line (linearization) to approximate a function value near $a$.
  • Find the normal line, requiring the negative reciprocal slope.
• Common mistakes
  • Using $f'(x)$ instead of $f'(a)$ as the slope in the tangent line equation.
  • Forgetting point-slope form and mixing up $a$ with $f(a)$.
  • Approximating at a point too far from $a$ and expecting high accuracy.

Putting It Together: Multi-Step Derivative Problems

Many exam questions test differentiation rules in combination with interpretation. The challenge is usually organizing the steps, using correct notation, and knowing what the derivative means in context.

Strategy for multi-step problems

1. Identify what is being asked: a derivative function, a derivative value, a slope, or a line.
2. Choose the right rule(s): linearity, power, product, quotient, trig, and common memory derivatives.
3. If evaluating at a point, compute the derivative first, then substitute.
4. In context, interpret with units and meaning (increasing/decreasing, rate, etc.).

Example: slope and tangent line with a mixed function

Let

$f(x)=(x^2+1)\sin(x)$

Find the equation of the tangent line at $x=0$.

Differentiate using the product rule. Let $u(x)=x^2+1$ and $v(x)=\sin(x)$.

$u'(x)=2x$

$v'(x)=\cos(x)$

So

$f'(x)=2x\sin(x)+(x^2+1)\cos(x)$

Evaluate at $x=0$:

$f(0)=(0^2+1)\sin(0)=0$

$f'(0)=2(0)\sin(0)+(0^2+1)\cos(0)=1$

Tangent line:

$y-0=1(x-0)$

So

$y=x$

Example: quotient derivative and evaluation

Let

$g(x)=\frac{\cos(x)}{x^2+1}$

Find $g'(0)$.

Use the quotient rule with numerator $f(x)=\cos(x)$ and denominator $h(x)=x^2+1$.

$f'(x)=-\sin(x)$

$h'(x)=2x$

So

$g'(x)=\frac{(-\sin(x))(x^2+1)-\cos(x)(2x)}{(x^2+1)^2}$

Evaluate at $0$:

$g'(0)=\frac{(-\sin(0))(1)-\cos(0)(0)}{1}=0$

Even though the derivative expression looks complicated, evaluating at a special point can simplify dramatically.

Exam Focus

• Typical question patterns
  • Combine rules (product and quotient with trig) and then evaluate at a specific $x$.
  • Given derivative information, build a tangent line and use it for an estimate.
  • Explain the meaning of a computed derivative in words (rate and units).
• Common mistakes
  • Substituting the point value before differentiating (unless you’re explicitly using the limit definition at that point).
  • Algebra errors when simplifying product and quotient derivatives (especially missing parentheses).
  • Giving a numerical answer without stating what it represents in a word problem.