15. Limits & Derivatives (Intro Calculus)
CollegeCalculus is the mathematics of change. The derivative measures an instantaneous rate — the slope of a curve at a single point — built on the foundational idea of a limit.
Limits and the derivative
A limit describes the value a function approaches as its input approaches some point, whether or not it is defined there. We write limx→a f(x) = L. The derivative is defined as a limit — the slope of the tangent line:
This is the limit of the average rate of change over an interval as the interval shrinks to zero. The essential rules:
Constant: d/dx [c] = 0
Sum: (f + g)′ = f′ + g′
Product: (fg)′ = f′g + fg′
Chain: d/dx [f(g(x))] = f′(g(x))·g′(x)
Where f′(x) = 0, the curve is momentarily flat — a candidate maximum or minimum.
Slope at a point, from slopes between points
Slope between two points is easy: rise over run. But what is the slope at a single point, where run is zero and division breaks? The derivative's trick is to compute the slope between x and a nearby point x + h, then slide that second point closer and closer. As h → 0, the "chord" pivots into the "tangent," and the average rate becomes the instantaneous rate.
This is why calculus is so powerful: velocity is the derivative of position, acceleration the derivative of velocity, marginal cost the derivative of cost. Any time you ask "how fast is this changing right now," you are asking for a derivative. The power rule is the workhorse — it says the slope of x² is 2x, matching the fact that a parabola gets steeper as you move out.
Worked example: derivative from first principles, then a maximum
Differentiate f(x) = x² using the limit definition.
- f(x + h) = (x + h)² = x² + 2xh + h².
- Difference quotient: [(x² + 2xh + h²) − x²]/h = (2xh + h²)/h = 2x + h.
- Let h → 0: f′(x) = 2x — matching the power rule.
A farmer has 40 m of fence for a rectangular pen against a wall (only three sides fenced). Maximise the area.
- Let the two sides perpendicular to the wall be x; the parallel side is 40 − 2x. Area A = x(40 − 2x) = 40x − 2x².
- Differentiate: A′ = 40 − 4x. Set to zero: x = 10.
- Then the parallel side is 40 − 20 = 20, giving A = 10 × 20 = 200 m². (Second derivative A″ = −4 < 0 confirms a maximum.)
Test your understanding
Differentiate y = 3x⁴ − 5x² + 7.
Apply the power rule term by term: d/dx[3x⁴] = 12x³, d/dx[−5x²] = −10x, and the constant 7 differentiates to 0. So y′ = 12x³ − 10x.
What is the physical meaning of the derivative of a position function?
It is velocity — the instantaneous rate of change of position with respect to time. If s(t) gives position, then s′(t) gives how fast and in which direction the object is moving at that exact instant. Differentiate again and you get acceleration.
Evaluate lim(x→0) (sin x)/x.
The limit is 1 (with x in radians). Although substituting 0 gives the indeterminate 0/0, for small angles sin x ≈ x, so the ratio approaches 1. This limit is exactly why the derivative of sin x is cos x, and it is a reason radians are the natural angle unit.
Use the chain rule to differentiate y = (2x + 1)⁵.
Outer function is "something to the 5th," inner is 2x + 1. Chain rule: y′ = 5(2x + 1)⁴ × d/dx(2x + 1) = 5(2x + 1)⁴ × 2 = 10(2x + 1)⁴. Differentiate the outside, keep the inside, then multiply by the inside's derivative.
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The process, step by step
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1Differentiate f(x) = x^2 from first principles.
- [(x+h)^2 - x^2]/h = (2xh + h^2)/h
- Simplify to 2x + h
- Limit as h to 0 gives 2x
Example 2Find the maximum of f(x) = -x^2 + 4x.
- f'(x) = -2x + 4
- Set -2x + 4 = 0 so x = 2
- f(2) = -4 + 8 = 4
Now you try
Work each one out first, then tap to reveal the worked answer.