Difficulty: Intermediate | Prerequisites: Average rate of change, basic algebra (FOIL, fraction arithmetic, factoring)
The limit definition of the derivative is where calculus properly begins. You already know how to find the average rate of change between two points. The derivative asks: what happens when those two points merge into one? This topic requires you to be comfortable with algebraic manipulation, especially expanding polynomials (FOIL), combining fractions with different denominators, and factoring. If fraction algebra feels shaky, review that before diving in.
The derivative of a function at a point is the limit of the difference quotient as h approaches zero: f'(x) = lim(h→0) [f(x+h) – f(x)] / h. To use this definition, you substitute, expand, simplify until h cancels from the denominator, then set h = 0. The algebra is the hard part; the calculus concept itself is straightforward.
Derivative, f'(x)
The instantaneous rate of change of f at x. Formally defined as f'(x) = lim(h→0) [f(x+h) – f(x)] / h, provided the limit exists. Think of it as the slope of the curve at a single point, rather than between two points.
Limit definition of the derivative
The formula f'(x) = lim(h→0) [f(x+h) – f(x)] / h. This is the "long way" of finding a derivative, before shortcut rules like the power rule. In simple terms, you let the two points of a secant line slide together until you have a tangent line.
Difference quotient
The expression [f(x+h) – f(x)] / h before you take the limit. It is the average rate of change over the interval from x to x+h. The derivative is what this expression becomes when h reaches zero.
FOIL (First, Outside, Inside, Last)
A method for multiplying two binomials: (a + b)(c + d) = ac + ad + bc + bd. You will use this constantly when expanding expressions like (x + h)² or (x + h)³ in limit-definition problems.
Tangent line
A line that touches the curve at exactly one point (locally) and has slope equal to the derivative at that point. The limit definition is the formal justification for why the tangent line has that slope.
Write the definition: f'(x) = lim(h→0) [f(x+h) – f(x)] / h.
Compute f(x+h) by replacing every x in the function with (x+h).
Subtract f(x) from f(x+h) in the numerator.
Simplify the numerator. The goal is to make every term contain a factor of h so that h cancels with the h in the denominator.
Cancel h from numerator and denominator.
Take the limit by setting h = 0 in whatever remains.
The algebra in step 4 is where most of the work happens. The type of algebra depends on the function.
Example: f(x) = 7x², find f'(1)
f(x+h) = 7(1+h)². Use FOIL: 7(1+h)(1+h) = 7(1 + 2h + h²) = 7 + 14h + 7h².
f(x) = 7(1)² = 7.
Numerator: (7 + 14h + 7h²) – 7 = 14h + 7h².
Divide by h: (14h + 7h²)/h = h(14 + 7h)/h = 14 + 7h.
Limit as h→0: 14 + 7(0) = 14.
So f'(1) = 14.
Example: f(x) = 9x³, find f'(x)
f(x+h) = 9(x+h)³. Expand (x+h)³ step by step: first (x+h)² = x² + 2xh + h², then multiply by (x+h) to get x³ + 3x²h + 3xh² + h³.
So f(x+h) = 9x³ + 27x²h + 27xh² + 9h³.
Numerator: 9x³ + 27x²h + 27xh² + 9h³ – 9x³ = 27x²h + 27xh² + 9h³.
Factor out h: h(27x² + 27xh + 9h²). Cancel with denominator.
Limit: 27x² + 27x(0) + 9(0)² = 27x².
So f'(x) = 27x².
Example: f(x) = 5x³, find f'(x)
Same expansion process. f(x+h) = 5(x+h)³ = 5x³ + 15x²h + 15xh² + 5h³.
Numerator after subtracting: 15x²h + 15xh² + 5h³.
Factor h, cancel, take limit: f'(x) = 15x².
Example: f(x) = 1/(9+x), find f'(x)
f(x+h) = 1/(9+x+h), f(x) = 1/(9+x).
Numerator of the difference quotient: [1/(9+x+h) – 1/(9+x)] / h.
Combine the fractions in the numerator by finding a common denominator: (9+x)(9+x+h). The numerator becomes [(9+x) – (9+x+h)] = –h.
The expression is now [–h] / [h(9+x+h)(9+x)].
Cancel h: –1 / [(9+x+h)(9+x)].
Limit as h→0: –1 / (9+x)².
So f'(x) = –1/(9+x)².
Example: f(x) = 8/(8+x), find f'(x)
Same approach. f(x+h) = 8/(8+x+h).
Combine: [8/(8+x+h) – 8/(8+x)] / h. Common denominator (8+x)(8+x+h). Numerator: 8(8+x) – 8(8+x+h) = –8h.
Simplify: –8h / [h(8+x+h)(8+x)]. Cancel h.
Limit: –8/(8+x)².
So f'(x) = –8/(8+x)².
Example: g(t) = 4/t², find g'(t)
g(t+h) = 4/(t+h)², g(t) = 4/t².
Combine: [4/(t+h)² – 4/t²] / h. Common denominator t²(t+h)². Numerator: 4t² – 4(t+h)² = 4t² – 4(t² + 2th + h²) = –8th – 4h².
Factor: h(–8t – 4h) / [h · t²(t+h)²]. Cancel h.
Limit: (–8t – 0) / [t² · t²] = –8t/t⁴ = –8/t³.
So g'(t) = –8/t³.
Expanding (x+h)² and (x+h)³: FOIL for squares; for cubes, FOIL the square first, then multiply by (x+h) again.
Combining fractions: find a common denominator, subtract numerators, and look for cancellation.
Factoring out h: after simplifying the numerator, every remaining term should contain h as a factor. If it does not, recheck your algebra.
Cancelling h: this is the step that makes the limit possible. If h does not cancel, the limit does not exist in finite form, which usually means an error.
The limit definition of the derivative:
f'(x) = lim(h→0) [f(x+h) – f(x)] / h
Useful algebraic identities for these problems:
(a + b)² = a² + 2ab + b²
(a + b)³ = a³ + 3a²b + 3ab² + b³
To combine fractions: a/b – c/d = (ad – bc) / bd
The derivative gives you the instantaneous rate of change, which is how engineers measure things like velocity at a precise moment (rather than average speed over a trip), how economists compute marginal cost (the cost of producing one more unit), and how physicists describe acceleration. Every time you see "rate" in a real-world context, a derivative is lurking behind it.
Students often forget to replace every x with (x+h) when computing f(x+h). If your function is 9x³, then f(x+h) = 9(x+h)³, not 9x³ + h or 9(x³ + h).
A common algebra error is expanding (x+h)² as x² + h², forgetting the cross term 2xh. This mistake will propagate through the entire problem.
Students sometimes try to take the limit before cancelling h from the denominator. If you set h = 0 while h is still in the denominator, you get 0/0, which is undefined. The algebra must come first.
When working with fractions, students frequently forget to distribute the negative sign when subtracting f(x). For example, 8(8+x) – 8(8+x+h) requires distributing the –8 through the entire second expression.
⚠️ Limit-definition problems are nearly guaranteed on the first major exam. Professors use them to test algebraic fluency and conceptual understanding before introducing shortcut rules.
⚠️ You will likely see at least one polynomial (use FOIL/expand) and one rational function (combine fractions). Know both types.
⚠️ Some exams ask you to find f'(x) in general, while others ask for f'(a) at a specific value. For f'(a), you substitute the number for x at the start, which sometimes makes the algebra simpler.
⚠️ Later in the course, you will use shortcut rules (power rule, quotient rule). Professors still test the limit definition separately because it demonstrates understanding of what a derivative actually is.
True or false: To find f(x+h) when f(x) = 3x², you write 3(x+h)², not 3x² + h. (True)
Fill in the blank: (x+h)² expands to x² + ___ + h². (2xh)
True or false: You can set h = 0 in the difference quotient before simplifying. (False. You must cancel h first.)
Fill in the blank: If f(x) = 1/(9+x), then f'(x) = ___. (–1/(9+x)²)
True or false: The derivative of 7x² at x = 1 is 7. (False. It is 14.)
Q: Use the limit definition to find f'(x) if f(x) = 4x².
A: f(x+h) = 4(x+h)² = 4x² + 8xh + 4h². Numerator: 8xh + 4h². Factor: h(8x + 4h). Cancel h, take limit: f'(x) = 8x.
Q: Use the limit definition to find f'(x) if f(x) = 1/(5+x).
A: Combine [1/(5+x+h) – 1/(5+x)] / h. Common denominator gives numerator –h. Cancel h, take limit: f'(x) = –1/(5+x)².
Q: Find f'(2) for f(x) = 3x³ using the limit definition.
A: f(2+h) = 3(2+h)³ = 3(8 + 12h + 6h² + h³) = 24 + 36h + 18h² + 3h³. Subtract f(2) = 24. Numerator: 36h + 18h² + 3h³. Factor h, cancel, set h = 0: f'(2) = 36.
Q: Why does every term in the numerator contain h after you subtract f(x) from f(x+h)?
A: Because f(x) cancels with the constant terms that appear when you expand f(x+h). The terms left over are all the ones that were generated by the +h, so they all have at least one factor of h. This is what makes the cancellation with the denominator possible.
Q: Use the limit definition to find g'(t) if g(t) = 6/t.
A: g(t+h) = 6/(t+h). Combine: [6/(t+h) – 6/t] / h. Numerator: 6t – 6(t+h) = –6h. Expression: –6h / [h · t(t+h)]. Cancel h: –6 / [t(t+h)]. Limit: –6/t².
This topic grows directly out of the average rate of change. The limit definition is what happens when you shrink the interval of the average rate of change to zero width. Once you are comfortable with the limit definition, you will learn shortcut rules (power rule, product rule, quotient rule, chain rule) that give the same answers far more quickly, but those rules are derived from, and justified by, this definition.
The derivative also connects to the equation of the tangent line: once you know f'(a), the tangent line at x = a is y = f(a) + f'(a)(x – a). That connects forward to linear approximation and differentials.
limit definition of derivative, derivative from first principles, difference quotient, f prime, f'(x), instantaneous rate of change, FOIL method, expanding polynomials, combining fractions, derivative of x squared, derivative of x cubed, derivative of 1 over x, derivative of rational function, brief calculus, calculus derivative problems, limit h approaches zero, tangent line slope, first principles derivative