A derivative measures how fast something is changing at one instant. If a car’s position is a function of time, the derivative is its speed at a given moment, the number on the speedometer. On a graph, the derivative at a point is the slope of the curve there.
From average rate to instant rate
You already know the average rate of change: the slope between two points.
Example Find the average rate of change of f(x) = x² from x = 1 to x = 3
(f(3) − f(1)) / (3 − 1) = (9 − 1)/2 = 4.
That’s the slope of the straight line through two points on the curve. Slide the second point closer and closer to the first, and the slope approaches the slope of the curve at that single point. That limit is the derivative.
The limit definition
f′(x) = lim (h → 0) of [f(x + h) − f(x)] / h
Example Use the definition to find the derivative of f(x) = x²
- f(x + h) = (x + h)² = x² + 2xh + h².
- Subtract f(x): 2xh + h².
- Divide by h: 2x + h.
- Let h go to 0: f′(x) = 2x.
So the slope of y = x² at x = 3 is 6, and at x = −1 it is −2. In practice, nobody uses the definition every time. The rules below are shortcuts proven from it.
The basic rules
| Rule | Statement | Example |
|---|---|---|
| Constant | d/dx (c) = 0 | d/dx (7) = 0 |
| Power | d/dx (xn) = nxn−1 | d/dx (x⁵) = 5x⁴ |
| Constant multiple | d/dx (c·f) = c·f′ | d/dx (3x²) = 6x |
| Sum and difference | d/dx (f ± g) = f′ ± g′ | Differentiate term by term |
Example Differentiate 5x⁴ − 3x² + 7x − 2
Term by term: 20x³ − 6x + 7. The constant −2 disappears.
The power rule works for any exponent, so rewrite roots and fractions as powers first.
Example Differentiate √x and 1/x²
- √x = x1/2, so its derivative is ½x−1/2 = 1/(2√x).
- 1/x² = x−2, so its derivative is −2x−3 = −2/x³.
Product rule
(fg)′ = f′g + fg′
Example Differentiate x² sin x
f = x² and g = sin x. So f′g + fg′ = 2x sin x + x² cos x.
The derivative of a product is not the product of derivatives
d/dx (x² · x³) is not 2x · 3x² = 6x³. It’s d/dx (x⁵) = 5x⁴. Use the product rule, or simplify first.
Quotient rule
(f/g)′ = (f′g − fg′) / g²
A common way to remember the order: “low d-high minus high d-low, over low squared.”
Example Differentiate x/(x + 1)
f′g − fg′ = 1·(x + 1) − x·1 = 1. Divide by (x + 1)²: the derivative is 1/(x + 1)².
Chain rule
For a function inside another function, take the derivative of the outside, leave the inside alone, then multiply by the derivative of the inside.
d/dx f(g(x)) = f′(g(x)) · g′(x)
Example Differentiate (3x + 1)⁵
Outside: 5(3x + 1)⁴. Inside derivative: 3. Answer: 15(3x + 1)⁴.
Example Differentiate sin(x²) and e2x
- sin(x²): cos(x²) · 2x = 2x cos(x²).
- e2x: e2x · 2 = 2e2x.
Derivatives to know by heart
| f(x) | f′(x) |
|---|---|
| sin x | cos x |
| cos x | −sin x |
| tan x | sec²x |
| ex | ex |
| ln x | 1/x |
| ax | ax ln a |
Tangent lines
The derivative gives the slope. A point on the curve gives the rest. Then use point-slope form.
Example Find the tangent line to f(x) = x³ at x = 2
- The point: f(2) = 8, so (2, 8).
- The slope: f′(x) = 3x², so f′(2) = 12.
- Point-slope: y − 8 = 12(x − 2), which is y = 12x − 16.
Where the derivative is zero, the tangent line is flat. Those points are where a graph can have a peak or a valley, which is how calculus finds maximums and minimums.
Motion
If s(t) is position, then s′(t) is velocity and the derivative of velocity is acceleration.
Example A ball is thrown upward with height s(t) = −16t² + 64t feet after t seconds. When does it reach its highest point, and how high is it?
- Velocity: v(t) = s′(t) = −32t + 64.
- At the top, the ball stops rising, so v(t) = 0: −32t + 64 = 0, giving t = 2 seconds.
- Height then: s(2) = −64 + 128 = 64 feet.
Problems to try
Differentiate 4x³ − x + 9.
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12x² − 1.
Differentiate 6√x.
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3/√x. 6 · ½x−1/2 = 3x−1/2.
Differentiate x²ex.
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2xex + x²ex, which factors to xex(x + 2).
Differentiate (2x − 5)³.
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6(2x − 5)². The chain rule gives 3(2x − 5)² · 2.
Differentiate ln(x² + 1).
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2x/(x² + 1). The outside derivative is 1/(x² + 1), times the inside derivative 2x.
Find the tangent line to f(x) = x² − 4x at x = 1.
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y = −2x − 1. f(1) = −3 and f′(1) = 2(1) − 4 = −2.
Where does f(x) = x³ − 12x have horizontal tangent lines?
Show answer
At x = 2 and x = −2, where f′(x) = 3x² − 12 = 0.
Practice
Polynomial Derivative and Integral CalculatorPrintable worksheet
Questions students ask
What’s the difference between f′(x), dy/dx and y′?
Nothing but notation. They all mean the derivative. dy/dx is handy because it reminds you it’s a change in y over a change in x.
Do I need to understand limits first?
A basic idea of limits helps you understand why the rules work. You can learn the rules and use them well before limits feel completely comfortable, which is how most courses run in practice.
What should I review before calculus?
Functions, exponents and slope do most of the work. See Functions Explained, Exponent Rules and Slope and Linear Equations.