Derivative Definition & Notation | AP Calculus Topic 2.2

1. The derivative as a function

Core concept: Topic 2.1 defined an instantaneous rate at a fixed point. Here we build the derivative function, use precise notation, and connect its values with tangent-line slopes. The domain of a derivative consists of inputs where the required finite limit exists; it need not equal the original function's domain.

THE DERIVATIVE FUNCTION

The derivative of a function $f$ is a function $f^{'}$ . Its value $f^{'} (x)$ is the instantaneous rate of change at an input where the defining limit exists as a finite real number.

Formal Definition (h-form):

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

Alternative Definition (x-form):

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

Function versus value: $f^{'}$ names the derivative function; $f^{'} (x)$ is its general value, while $f^{'} (a)$ is the number obtained at a fixed input $a$ . Either limit form can define a value at a point: replacing $x$ by $a$ in the input-change form also gives $f^{'} (a)$ .

Existence: At an interior input $a$ , the ordinary derivative $f^{'} (a)$ requires a finite two-sided difference-quotient limit. A divergent or unequal-sided limit does not give an ordinary derivative. Use only inputs in the function's domain; specify one-sided derivatives explicitly at endpoints.

1. Start with the function

$f (x) = x^{2} + 3 x$ assigns an output to each real input.

2. Build its derivative

The definition gives $f^{'} (x) = 2 x + 3$ , also defined for all real inputs in this example.

3. Evaluate a rate

$f^{'} (2) = 7$ , while $f (2) = 10$ . A slope and an output value are different quantities.

4. Interpret its units

Derivative units are $\frac{output units}{input units}$ . A temperature rate, for example, can be measured in $\frac{^{\circ} C}{m i n}$ .

The derivative domain may be smaller: $f (x) = \sqrt{x}$ has domain $[0, \infty)$ , but its finite ordinary derivative $f^{'} (x) = \frac{1}{2 \sqrt{x}}$ is defined only for $x > 0$ . At zero, the right-hand quotient diverges.

Review Topic 2.1: average and instantaneous rates and limit notation if the limiting process is unfamiliar.

2. Derivative notation and evaluation at a point

Equivalent notation needs the same dependent and independent variables

For a function $y = f (x)$ , the derivative can be written in multiple ways:

Notation	Name	Read As	Best Used For
$f^{'} (x)$	Lagrange (Prime)	"f prime of x"	General calculus, clean formulas
$\frac{d y}{d x}$	Leibniz	“dee y over dee x”	Related rates, implicit differentiation
$y^{'}$	Lagrange (Short)	"y prime"	Quick algebraic work
$\frac{d f}{d x}$	Leibniz (Function)	"dee f dee x"	Named function differentiated with respect to the stated variable
$\dot{y}$	Newton	"y dot"	Physics (time derivatives)
$D f$ or $D_{x} f$	Euler	"D of f"	Differential equations

Derivative at a Specific Point

When evaluating the derivative at a particular value $x = a$ :

f^{'} (a) or \frac{d y}{d x} |_{x = a} or y^{'} (a)

These denote the derivative evaluated at $x = a$ , with $y$ understood as the function $y (x) = f (x)$ . The evaluation bar applies to the entire derivative.

When to Use Which Notation:

$f^{'} (x)$ : Best for general derivative problems, applying derivative rules, graphing
$\frac{d y}{d x}$ : Perfect for related rates (emphasizes variables), implicit differentiation, chain rule
$y^{'}$ : Convenient shorthand when $y = f (x)$ and context is clear
$\dot{x}$ : Newton's dot denotes a time derivative when $x = x (t)$ ; velocity is $\dot{x} = \frac{d x}{d t}$ , and acceleration is $\dot{v} = \frac{d v}{d t} = \ddot{x}$ . It is not automatically a derivative with respect to $x$ .

Read the variable, not just the symbol. For $y = f (x)$ , $f^{'} (x) = y^{'} = \frac{d y}{d x}$ . Newton's dot means differentiation with respect to time; operator notation such as $D_{x} f$ needs its variable specified. These conventions do not make derivatives with different independent variables interchangeable.

$\frac{d y}{d x}$ is derivative notation, not a quotient of two finite changes. By contrast, $\frac{Δ y}{Δ x}$ is an average rate across a nonzero input change. Later differential and chain-rule manipulations require their own mathematical justification.

For a time-dependent position $s (t)$ , write $v (t) = s^{'} (t) = \frac{d s}{d t}$ , and at time three write $v (3) = s^{'} (3) = {\frac{d s}{d t} |}_{t = 3}$ . The evaluation bar specifies which derivative value is intended.

3. Tangent slope and the tangent-line equation

The Single Most Important Geometric Fact:

f^{'} (a) = m_{t a n}

This connection between derivatives (algebra/calculus) and tangent line slopes (geometry) is why derivatives are so powerful for understanding curves!

Equation of the Tangent Line

To find the equation of the line tangent to $y = f (x)$ at the point $(a, f (a))$ :

Find the slope: $m = f^{'} (a)$
Use point-slope form:
$y - f (a) = f^{'} (a) (x - a)$
Simplify if needed to slope-intercept form $y = m x + b$

Local meaning: A finite derivative determines the tangent's local slope. A tangent may cross the curve or intersect it again elsewhere. Near a differentiable point, the tangent gives the linear approximation $f (a + h) \approx f (a) + f^{'} (a) h$ , with an error that becomes negligible relative to $h$ as $h \to 0$ .

In the formula above, $m_{t a n}$ is the finite tangent slope at $(a, f (a))$ . A zero derivative gives a horizontal tangent; a vertical tangent has no finite derivative slope and cannot be inserted into the finite-slope formula.

When the derivative exists, an equivalent tangent formula is $y = f (a) + f^{'} (a) (x - a)$ . The tangent's slope uses $f^{'} (a)$ ; its anchor point uses $f (a)$ . Substituting one for the other is a common error.

4. Connect analytical, graphical, numerical and verbal forms

A derivative is one concept expressed in several ways. For $f (x) = x^{2} + 3 x$ at input two, the exact derivative is seven.

Analytical

$f^{'} (2) = lim_{h \to 0} \frac{f (2 + h) - f (2)}{h} = lim_{h \to 0} (7 + h) = 7$ .

Graphical

The graph passes through $(2, 10)$ . Its tangent has slope seven, so its equation is $y - 10 = 7 (x - 2)$ .

Numerical

The quotients below approach seven from both sides. A finite table supports an estimate; the algebraic limit supplies the justification.

Verbal

At input two, the output is increasing locally at seven output units per input unit. This does not claim that every later unit of input adds exactly seven.

Nearby exact quotients for the quadratic example
Input change	$\frac{f (2 + h) - f (2)}{h}$
$- 0.1$	$6.9$
$- 0.01$	$6.99$
$0.01$	$7.01$
$0.1$	$7.1$

See Topic 2.3: estimating derivatives from graphs and tables for estimates when a formula is unavailable.

5. Differentiability and continuity

DIFFERENTIABILITY

A function $f$ is differentiable at $x = a$ if $f^{'} (a)$ exists—that is, if the limit $lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$ exists and is finite.

A function is differentiable on an open interval if it has a finite derivative at every input in that interval. If endpoints of a closed interval are included under a convention, the corresponding one-sided derivative requirement must be stated separately.

KEY THEOREM: Differentiability Implies Continuity

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

Contrapositive: If $f$ is discontinuous at $x = a$ , then $f$ is NOT differentiable at $x = a$ .

Important: The converse is NOT true! A function can be continuous but NOT differentiable (e.g., $f (x) = | x |$ at $x = 0$ ).

Here is why differentiability forces continuity. For nonzero $h$ ,

f (a + h) - f (a) = h (\frac{f (a + h) - f (a)}{h}) .

If the quotient tends to the finite value $f^{'} (a)$ , the right-hand side tends to zero. Thus $lim_{h \to 0} f (a + h) = f (a)$ , which is continuity at the input. This proof does not reverse: continuity alone does not force a limiting slope.

Continue to Topic 2.4: differentiability and continuity for the full treatment. The visual checklist is useful, but the defining quotient—not an informal claim that a curve looks “smooth”—decides differentiability.

6. When a finite derivative does not exist

Four ways a finite $f' (a)$ can fail to exist

1. Discontinuity

If $f$ has a jump, removable, or infinite discontinuity at $x = a$ , then $f^{'} (a)$ does not exist.

Example: $f (x) = {\begin{matrix} x^{2} & x < 1 \\ x + 1 & x \geq 1 \end{matrix}$ is not differentiable at $x = 1$ (jump discontinuity)

2. Corner or Cusp (Sharp Point)

At a corner, the one-sided difference-quotient limits are finite but unequal. At a typical cusp, they diverge with opposite signs. Neither gives a single finite derivative.

Example: $f (x) = | x |$ at $x = 0$ has a corner

Left derivative: $- 1$ ; Right derivative: $+ 1$ → No derivative!

3. Vertical Tangent

A vertical tangent has no finite slope. For example, the difference quotient for $f (x) = \sqrt[3]{x}$ at zero is $\frac{\sqrt[3]{h}}{h} = \frac{1}{h^{\frac{2}{3}}}$ ; it tends to $+ \infty$ from both sides. The ordinary derivative is undefined.

Example: $f (x) = \sqrt[3]{x}$ at $x = 0$ has vertical tangent

4. Oscillating Behavior

If $f$ oscillates too wildly near $x = a$ , the difference quotient might not have a limit.

Corrected example: define $f (x) = x \sin (\frac{1}{x})$ for $x \neq 0$ , and $f (0) = 0$ . This extension is continuous at zero because $| f (x) | \leq | x |$ . But its quotient $\frac{f (h) - f (0)}{h} = \sin (\frac{1}{h})$ has no limit, so it is not differentiable at zero. Oscillation of the quotient is the decisive issue.

Quick Test for Differentiability:

Check continuity first: If not continuous, definitely not differentiable
Look for sharp points: If graph has corner/cusp, not differentiable
Check for vertical tangents: If tangent is vertical, not differentiable
Compare one-sided derivative limits: Compute $lim_{h \to 0^{-}} \frac{f (a + h) - f (a)}{h}$ and $lim_{h \to 0^{+}} \frac{f (a + h) - f (a)}{h}$ . Both must be finite and equal. These are limits of difference quotients, not merely limits of a formula for $f^{'}$ near the point.

7. Five worked examples with the limit definition

Example 1: build a derivative function

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

Solution:

Set up the limit:
$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$
Find $f (x + h)$ :
- $f (x + h) = (x + h)^{2} + 3 (x + h)$
- $= x^{2} + 2 x h + h^{2} + 3 x + 3 h$
Compute the numerator:
$f (x + h) - f (x) = (x^{2} + 2 x h + h^{2} + 3 x + 3 h) - (x^{2} + 3 x)$

$= 2 x h + h^{2} + 3 h$
Form the difference quotient:
$\frac{f (x + h) - f (x)}{h} = \frac{2 x h + h^{2} + 3 h}{h} = \frac{h (2 x + h + 3)}{h}$
Cancel $h$ and take the limit:
$f^{'} (x) = lim_{h \to 0} (2 x + h + 3) = 2 x + 3$

Answer: $f^{'} (x) = 2 x + 3$

Interpretation: At any input $x$ , the slope of the tangent line is $2 x + 3$

Example 2: evaluate a derivative at a point

Problem: Find $f^{'} (2)$ for $f (x) = x^{3}$ using the limit definition

Solution:

Set up with $a = 2$ :
$f^{'} (2) = lim_{h \to 0} \frac{f (2 + h) - f (2)}{h}$
Evaluate:
- $f (2 + h) = (2 + h)^{3} = 8 + 12 h + 6 h^{2} + h^{3}$
- $f (2) = 8$
Substitute:
$f^{'} (2) = lim_{h \to 0} \frac{(8 + 12 h + 6 h^{2} + h^{3}) - 8}{h} = lim_{h \to 0} \frac{12 h + 6 h^{2} + h^{3}}{h}$
Factor and cancel:
$= lim_{h \to 0} \frac{h (12 + 6 h + h^{2})}{h} = lim_{h \to 0} (12 + 6 h + h^{2})$
Evaluate the limit:
$f^{'} (2) = 12 + 0 + 0 = 12$

Answer: $f^{'} (2) = 12$

Meaning: At the point $(2, 8)$ , the slope of the tangent line is 12

Example 3: find a reciprocal tangent line

Problem: Find the equation of the line tangent to $f (x) = \frac{1}{x}$ at $x = 2$

Solution:

Find the point: $f (2) = \frac{1}{2}$ , so point is $(2, \frac{1}{2})$
Find the slope using limit definition:
$f^{'} (2) = lim_{h \to 0} \frac{f (2 + h) - f (2)}{h} = lim_{h \to 0} \frac{\frac{1}{2 + h} - \frac{1}{2}}{h}$
Common denominator in numerator:
$= lim_{h \to 0} \frac{\frac{2 - (2 + h)}{2 (2 + h)}}{h} = lim_{h \to 0} \frac{\frac{- h}{2 (2 + h)}}{h}$
Simplify:
$= lim_{h \to 0} \frac{- h}{2 (2 + h) h} = lim_{h \to 0} \frac{- 1}{2 (2 + h)} = \frac{- 1}{4}$
Use point-slope form:
$y - \frac{1}{2} = - \frac{1}{4} (x - 2)$
Simplify:
$y = - \frac{1}{4} x + \frac{1}{2} + \frac{1}{2} = - \frac{1}{4} x + 1$

Answer: $y = - \frac{1}{4} x + 1$

Example 4: compare one-sided derivatives

Problem: Determine if $f (x) = | x - 1 |$ is differentiable at $x = 1$

Solution:

Check continuity: $f (1) = 0$ , $lim_{x \to 1} f (x) = 0$ → Continuous
Check left-hand derivative:
- For $x < 1$ : $f (x) = - (x - 1) = 1 - x$
- $f_{-}^{'} (1) = lim_{h \to 0^{-}} \frac{f (1 + h) - f (1)}{h} = lim_{h \to 0^{-}} \frac{- h}{h} = - 1$
Check right-hand derivative:
- For $x > 1$ : $f (x) = x - 1$
- $f_{+}^{'} (1) = lim_{h \to 0^{+}} \frac{f (1 + h) - f (1)}{h} = lim_{h \to 0^{+}} \frac{h}{h} = 1$
Compare: $f_{-}^{'} (1) = - 1 \neq 1 = f_{+}^{'} (1)$

Conclusion: NOT differentiable at $x = 1$ because the one-sided derivatives don't match (corner point)

Example 5: use Leibniz notation

Problem: If $y = x^{2} - 4 x + 3$ , find $\frac{d y}{d x}$ using the limit definition

Solution:

Set up:
$\frac{d y}{d x} = lim_{h \to 0} \frac{y (x + h) - y (x)}{h}$
Expand:
- $y (x + h) = (x + h)^{2} - 4 (x + h) + 3$
- $= x^{2} + 2 x h + h^{2} - 4 x - 4 h + 3$
Subtract:
$y (x + h) - y (x) = 2 x h + h^{2} - 4 h$
Divide and simplify:
$\frac{d y}{d x} = lim_{h \to 0} \frac{h (2 x + h - 4)}{h} = lim_{h \to 0} (2 x + h - 4) = 2 x - 4$

Answer: $\frac{d y}{d x} = 2 x - 4$ (same as $y^{'} = 2 x - 4$ or $f^{'} (x) = 2 x - 4$ )

Domain and notation checks for the preserved examples

Examples 1, 2 and 5 use polynomials, so their derivative formulas are valid for every real input. In Example 3, the reciprocal requires $x \neq 0$ , and its quotient at two requires $h \neq 0$ and $h \neq - 2$ before cancellation. In Example 4, $f_{-}^{'} (1)$ and $f_{+}^{'} (1)$ denote the left and right derivative limits at the base point, respectively.

8. Higher-order derivative notation

Notation for Higher Derivatives

Second Derivative (derivative of the derivative):

f^{″} (x) or \frac{d^{2} y}{d x^{2}} or y^{″}

Third Derivative:

f^{‴} (x) or \frac{d^{3} y}{d x^{3}} or y^{‴}

nth Derivative:

f^{(n)} (x) or \frac{d^{n} y}{d x^{n}}

Higher-order notation: $f^{(n)} (x)$ means the nth derivative, not the nth power. Parenthesized notation such as $f^{(4)} (x)$ is clearer than a long string of primes. The expressions $\frac{d^{2} y}{d x^{2}}$ and ${(\frac{d y}{d x})}^{2}$ mean different things.

The second derivative measures the rate of change of the first derivative. If $s$ is position and $t$ is time, $s^{'} (t) = v (t)$ and $s^{″} (t) = v^{'} (t) = a (t)$ . Position in meters and time in seconds give velocity units $\frac{m}{s}$ and acceleration units $\frac{m}{s^{2}}$ . Higher derivatives exist only where the successive derivatives exist.

9. Methods, notation choices and common mistakes

Essential Tips

Simplify the quotient on its allowed domain: Factor for polynomials, combine fractions for rational functions, or multiply by a conjugate for radicals. Cancel common factors only when nonzero.
Do not substitute $h = 0$ into the original quotient: It is undefined there. A $\frac{0}{0}$ form signals that the limit requires further analysis; it is not a value or proof that the derivative fails.
For the same function and independent variable: $f^{'} (x) = \frac{d y}{d x} = y^{'}$ — equivalent notation in this context
Differentiable implies continuous: But continuous doesn't guarantee differentiable!
Check one-sided derivatives: At potential problem points (corners, cusps)
Tangent line needs two things: Point and slope (derivative)
For notation choice: Use what makes the problem clearest

The 5-Step Limit Definition Method

A five-step method for evaluating the derivative definition:

SET UP: Write $f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$
EXPAND: Find $f (x + h)$ by substituting
SUBTRACT: Compute $f (x + h) - f (x)$ and simplify
SIMPLIFY: For a polynomial, factor $h$ from the entire numerator; for a reciprocal or radical, combine fractions or rationalize as needed.
EVALUATE: Cancel a nonzero common factor when justified, take the limit, and state the derivative's domain. Preserve the limit notation until the limit has been evaluated.

Memory aid: "SESFC" (Set, Expand, Subtract, Factor, Cancel)

Quick Notation Guide

If you see...	Use notation...
“Find $f'$ ” or "named function"	$f^{'} (x)$ or $f^{'} (a)$
"Related rates" or "implicit"	$\frac{d y}{d x}$ , $\frac{d r}{d t}$
“ $y = f (x)$ ” (simple equation)	$y^{'}$ or $\frac{d y}{d x}$
Physics/velocity problems	$\dot{x}$ , $\dot{v}$ (Newton)
Differential equations	$D y$ or $\frac{d y}{d x}$

Common Mistakes to Avoid

Mistake 1: Substituting $h = 0$ into an undefined original difference quotient instead of evaluating its limit.
Mistake 2: Forgetting to distribute when expanding $f (x + h)$
Mistake 3: Not factoring out $h$ correctly—must factor from entire numerator
Mistake 4: Assuming continuity guarantees differentiability. For example, $f (x) = | x |$ is continuous at zero but has no derivative there.
Mistake 5: Confusing $f^{'} (a)$ (number) with $f^{'} (x)$ (function)
Mistake 6: Losing the independent variable or evaluation point. Prefer ${\frac{d y}{d x} |}_{x = a}$ for a value at a point. Writing a derivative function with an explicit argument can be valid if that function notation has been defined; it is not universally forbidden.
Mistake 7: Forgetting the limit symbol entirely when using definition
Mistake 8: Not checking for corners/cusps when testing differentiability

10. Six practice problems with complete solutions

For Problems 1–4, use the limit definition to find the derivative function. The last two problems apply tangent-line and differentiability reasoning.

1. A linear function

$f (x) = 5 x - 2$ .

Check the calculation and domain

\frac{f (x + h) - f (x)}{h} = \frac{5 (x + h) - 2 - (5 x - 2)}{h} = \frac{5 h}{h} = 5, h \neq 0 .

f^{'} (x) = lim_{h \to 0} 5 = 5 .

The line's slope is constant; the derivative domain is all real inputs.

2. A quadratic function

$f (x) = 4 x^{2}$ .

Check the calculation and domain

\frac{4 (x + h)^{2} - 4 x^{2}}{h} = \frac{8 x h + 4 h^{2}}{h} = 8 x + 4 h, h \neq 0 .

f^{'} (x) = lim_{h \to 0} (8 x + 4 h) = 8 x .

The derivative domain is all real inputs. The rate varies with the input.

3. A shifted reciprocal

$f (x) = \frac{1}{x + 1}$ .

Check the calculation and domain

\frac{f (x + h) - f (x)}{h} = \frac{\frac{1}{x + h + 1} - \frac{1}{x + 1}}{h} .

= \frac{- h}{h (x + h + 1) (x + 1)} = \frac{- 1}{(x + h + 1) (x + 1)} .

f^{'} (x) = lim_{h \to 0} \frac{- 1}{(x + h + 1) (x + 1)} = - \frac{1}{(x + 1)^{2}} .

Require $x \neq - 1$ , $h \neq 0$ , and $x + h \neq - 1$ while simplifying. The derivative domain is $x \neq - 1$ , the same as the original domain here.

4. A square-root function

$f (x) = \sqrt{x}$ .

Check the calculation and domain

For a positive fixed input and allowed nonzero changes, multiply by the conjugate:

\frac{\sqrt{x + h} - \sqrt{x}}{h} = \frac{(x + h) - x}{h (\sqrt{x + h} + \sqrt{x})} .

= \frac{1}{\sqrt{x + h} + \sqrt{x}}, h \neq 0 .

f^{'} (x) = lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}} = \frac{1}{2 \sqrt{x}}, x > 0 .

The function includes zero in its domain, but the finite derivative does not: at zero the right-hand quotient $\frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}}$ tends to $+ \infty$ .

5. A tangent-line equation

Find the tangent to $y = x^{2}$ at $x = 3$ .

Check the point, slope and line

f (3) = 9, f^{'} (3) = lim_{h \to 0} \frac{(3 + h)^{2} - 9}{h} = lim_{h \to 0} (6 + h) = 6 .

y - 9 = 6 (x - 3), y = 6 x - 9 .

The line passes through $(3, 9)$ , and its slope equals the derivative there.

6. A cusp at zero

Is $f (x) = x^{\frac{2}{3}}$ differentiable at $x = 0$ ?

Check the one-sided quotient limits

Interpret the real power as $(\sqrt[3]{x})^{2}$ . The function is continuous at zero. For nonzero changes,

\frac{f (h) - f (0)}{h} = \frac{h^{\frac{2}{3}}}{h} = \frac{1}{\sqrt[3]{h}} .

lim_{h \to 0^{-}} \frac{1}{\sqrt[3]{h}} = - \infty, lim_{h \to 0^{+}} \frac{1}{\sqrt[3]{h}} = + \infty .

The opposite unbounded one-sided slopes form a cusp. There is no finite derivative at zero. This differs from the cube-root vertical-tangent example, whose quotients diverge positively from both sides.

11. AP-style reasoning and written solutions

What the AP® Exam Expects:

Show the limit setup: Write $f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ explicitly
Justify the calculation: When the definition is requested, show the limit setup and enough valid algebra to establish it. A later derivative-rule result alone does not demonstrate use of the definition.
Use proper notation: Be consistent with $f^{'} (x)$ , $\frac{d y}{d x}$ , or $y^{'}$
State the derivative value: "Therefore, $f^{'} (x) = . . .$ " as your conclusion
For tangent lines: Show finding the slope AND using point-slope form
Justify non-differentiability: Explain WHY (corner, discontinuity, vertical tangent)
Check one-sided derivatives: When asked about differentiability at a point

Common FRQ Formats:

"Use the definition of derivative to find $f' (x)$ "
"Find the equation of the line tangent to $f$ at $x = a$ "
"Show that f is not differentiable at $x = a$ "
"Using correct notation, express the instantaneous rate of change"
"Find $f' (3)$ using the limit definition"
"Determine if f is differentiable at $x = 2$ . Justify your answer."

12. Quick reference

Concept	Key Formula/Fact
Derivative Definition (h-form)	$f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$
Derivative Definition (x-form)	$f^{'} (a) = lim_{x \to a} \frac{f (x) - f (a)}{x - a}$
Geometric Meaning	Slope of tangent line at point
Tangent Line Equation	$y - f (a) = f^{'} (a) (x - a)$
Differentiable → Continuous	Always true
Continuous → Differentiable	Not always true: $f (x) = \| x \|$ is a counterexample at zero.
Equivalent Notations	$f^{'} (x) = \frac{d y}{d x} = y^{'} = D f$

13. Connections and next study steps

Topic 2.2 connects to:

Topic 2.1: Extends instantaneous rate from a point to a function
Topics 2.5–2.10: Power, sum, trigonometric, product and quotient rules provide shortcuts for derivative limits.
Unit 3: Composite, implicit and inverse differentiation use the derivative definition and notation.
Unit 4: Contextual rates, motion and related rates use derivatives and meaningful units.
Unit 5: Analytical applications use first and second derivatives for behavior, concavity and optimization. Acceleration is a second derivative of position in motion applications.
Unit 6: Integration and accumulation connect to differentiation through the Fundamental Theorem of Calculus.
All of calculus: This is THE foundational definition!

Keep three ideas distinct: the original function gives output values, its derivative function gives local rates where the finite limit exists, and a derivative value supplies a tangent slope at a specified point. Use complete limit notation, valid algebra, appropriate domains and clearly defined variables.

← Topic 2.1: average and instantaneous rates Next: Topic 2.3 — estimating derivatives →All Unit 2 study notes →Review algebraic limit techniques →

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