Defining the Derivative of a Function and Using Derivative Notation: Study Guide (Calculus AB 2.2)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so limits are written as lim (h → 0) g(h), meaning “the limit as h approaches 0 of g(h)”. The derivative is written with a prime, f′(x), read “f prime of x”. The value of dy/dx at a particular point, say x = 3, is written here as dy/dx at x = 3. On paper you will often see a vertical bar after dy/dx with “x = 3” written small at its foot; both mean the same.

From a number to a function

In Topic 2.1 you found the derivative at one point a:

f′(a) = lim (h → 0) (f(a + h) − f(a))/h

You can do this at a = 1, then at a = 2, then at a = 3, and so on. Each time you get a number. Instead of repeating the work, keep the point as a variable, x:

Definition. The derivative of f is the function f′ whose value at x is f′(x) = lim (h → 0) (f(x + h) − f(x))/h, provided this limit exists.

Inside the limit, x is held fixed while h → 0. Only after the limit is taken do you let x vary. The result is a new function. Feed it any input a where the limit exists, and it returns f′(a), the slope of the graph of f at that input.

A first example. Let f(x) = x²/2. Then

f(x + h) − f(x) = (x² + 2xh + h²)/2 − x²/2 = xh + h²/2.

Divide by h (h ≠ 0) to get x + h/2. Let h → 0: f′(x) = x.

One calculation now answers infinitely many questions. The slope of y = x²/2 is −2 at x = −2, 0 at x = 0, 1 at x = 1 and 3 at x = 3. (Topic 2.1, Figure 1 showed the slope 1 at x = 1 on this same curve.)

Where f′ exists. f′(x) is defined only at inputs where the limit exists. This can be fewer inputs than f has. For example, √x is defined at x = 0, but its derivative from the definition, 1/(2√x), exists only for x > 0. Topic 2.4 looks at such points in detail.

Derivative notation

For y = f(x), these all name the derivative:

Notation	Read as	Notes
f′(x)	“f prime of x”	Names the function f. Value at a point: f′(3)
y′	“y prime”	Short; use when the function is called y
dy/dx	“d y d x”, or “the derivative of y with respect to x”	Shows both variables. Value at a point: dy/dx at x = 3
d/dx (f(x))	“the derivative of f(x) with respect to x”	Useful for an expression without a name, e.g. d/dx (x²/2) = x

Why dy/dx looks like a fraction. It comes from the average rate Δy/Δx, the change in y over the change in x. The “d” signals that the limit has been taken. In this unit, treat dy/dx as a single symbol.

Other letters. If s(t) is a position at time t, its derivative is s′(t) or ds/dt. The letters change; the meaning does not. The variable in the bottom (dt, dx) tells you the input you are differentiating with respect to.

Two common slips. f′(3) means “find f′(x), then put x = 3”. It does not mean “put x = 3 into f, then differentiate the number”: that would give 0, because a constant has slope 0. And y′ is not y: y is a height on the graph, y′ is a slope.

Worked example 1: a polynomial from the definition

Question. Let f(x) = 2x² − 3x + 1. Use the definition to find f′(x). Then find f′(0), f′(3/4) and f′(2).

1. Write f(x + h). Replace every x by (x + h): f(x + h) = 2(x + h)² − 3(x + h) + 1 = 2x² + 4xh + 2h² − 3x − 3h + 1.
2. Subtract f(x). The terms 2x², −3x and +1 cancel: f(x + h) − f(x) = 4xh + 2h² − 3h
3. Divide by h (h ≠ 0): (f(x + h) − f(x))/h = 4x + 2h − 3
4. Take the limit, holding x fixed: f′(x) = lim (h → 0) (4x + 2h − 3) = 4x − 3
5. Evaluate. f′(0) = −3, f′(3/4) = 0, f′(2) = 5.

Answer. f′(x) = 4x − 3; f′(0) = −3, f′(3/4) = 0, f′(2) = 5.

Interpretation. At x = 0 the graph of f falls with slope −3. At x = 3/4 the tangent line is horizontal. At x = 2 the graph rises with slope 5.

Check. Every term that does not contain h must cancel in step 2. If one survives, you have made an expansion error, because the top must be 0 when h = 0.

Worked example 2: a rational function

Question. Let f(x) = 3/(x + 1). Use the definition to find f′(x), and state where it exists.

1. Write the difference quotient. (f(x + h) − f(x))/h = [3/(x + h + 1) − 3/(x + 1)] / h
2. Combine the fractions on top over the common denominator (x + h + 1)(x + 1): 3(x + 1) − 3(x + h + 1) = 3x + 3 − 3x − 3h − 3 = −3h. So the top is −3h / ((x + h + 1)(x + 1)).
3. Divide by h and cancel (h ≠ 0): = −3 / ((x + h + 1)(x + 1))
4. Take the limit. As h → 0, x + h + 1 → x + 1: f′(x) = −3/(x + 1)²
5. Where it exists. The formula works for every x ≠ −1. At x = −1, f itself is undefined, so there is no derivative there either.

Answer. f′(x) = −3/(x + 1)², for x ≠ −1.

Interpretation. The numerator is negative and (x + 1)² is positive, so f′(x) < 0 everywhere it exists: the graph of f falls on each side of its asymptote. For example, f′(0) = −3, f′(2) = −1/3 and f′(−3) = −3/4.

The tangent line

f′(a) is the slope of the tangent line to y = f(x) at the point (a, f(a)). A line through a known point with a known slope has the point-slope equation

Tangent line at x = a: y − f(a) = f′(a)(x − a)

You need two numbers: the height f(a) and the slope f′(a).

Worked example 3: writing a tangent line

Question. Find an equation of the line tangent to the graph of y = 3/(x + 1) at x = 2.

1. Point. f(2) = 3/3 = 1. The point is (2, 1).
2. Slope. From Worked example 2, f′(x) = −3/(x + 1)². So f′(2) = −3/9 = −1/3.
3. Point-slope form. y − 1 = −(1/3)(x − 2)
4. Optional rearrangement. y = −x/3 + 5/3. Point-slope form is acceptable as a final answer unless a question asks for another form.

Answer. y − 1 = −(1/3)(x − 2), or y = (5 − x)/3.

Check. Near x = 2, the tangent line should be close to the curve. At x = 2.3 the line gives 0.9 and the curve gives 3/3.3 ≈ 0.909. Close, as expected.

Four representations of a derivative

A derivative can be given, or asked for, in any of four forms. Be ready to move between them.

Representation	What it looks like	Example from this page
Analytical	A formula for f′(x), or a limit	f′(x) = 4x − 3, from the definition
Numerical	A table of derivative values	For f(x) = 3/(x + 1): f′(0) = −3, f′(2) = −1/3
Graphical	The slope of the tangent line at each point of the graph of f	Figure 1: slope −1/3 at (2, 1)
Verbal	A sentence about the rate of change, with units	“At t = 5 minutes, the volume is decreasing at 2 litres per minute” means V′(5) = −2, if V(t) is in litres and t in minutes

Units. The units of f′(x) are the units of f divided by the units of x. If C(n) is a cost in dollars for n kilograms, then C′(n) is in dollars per kilogram. Topic 2.3 adds estimating derivatives from tables and graphs.

Common misconceptions

• “f′(x) and f′(a) are the same kind of thing.” f′(x) is a function; f′(a) is one number. A tangent line needs the number.
• Using f′(x) as the slope of a tangent line. “y − 3 = (4x − 3)(x − 2)” is not a line. Substitute x = a into f′ first.
• Using f(a) as the slope, or f′(a) as the height. The point is (a, f(a)); the slope is f′(a).
• Substituting before differentiating. For f(x) = 2x² − 3x + 1, f(2) = 3 is a constant. “Differentiating 3” gives 0, but f′(2) = 5.
• Expanding f(x) + h instead of f(x + h). Replace every x by (x + h), including inside powers and denominators.
• Splitting dy/dx. In this unit, dy/dx is one symbol, not d × y ÷ (d × x).
• “The derivative exists wherever the function does.” √x is defined at 0, but its derivative is not. And f′ never exists where f is undefined.
• “A tangent line touches the curve only once.” It touches the curve at the point of tangency, but it may cross the curve somewhere else. What defines it is the slope f′(a) at that point.

Where this leads

You now have the derivative as a function, three ways to write it and a way to find tangent lines. The next topic, Estimating Derivatives of a Function at a Point, estimates f′(a) from tables and graphs, and with technology. Topic 2.4 explains when the derivative fails to exist, and from Topic 2.5 onwards you meet the rules that let you skip the limit for familiar functions. For the previous topic, see Defining Average and Instantaneous Rates of Change at a Point. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.