Mathematics

Intro to derivatives

From the slope between two points to the slope at exactly one point.

What you need first

The average rate of change tells you how steep things are on average between two points. But what if you want to know how fast something changes in a single instant, for example your speed right now? For that you need the slope at a single point. This is exactly what the derivative delivers.

From secant to tangent

Take a fixed point on the graph and a second one next to it. The line through both is the , and you already know its slope. Now you slide the second point closer and closer to the first. The secant tilts as you do and turns into a line that touches the graph at that spot and has exactly the same slope there as the graph itself: the tangent.

Coordinate grid

y = 1x² + 1opens upwards

Move the second point toward the first and watch the secant become a tangent.

Instantaneous rate of change

The closer the second point comes, the smaller the gap $h$ between the positions gets. The difference in height divided by $h$ is called the , and it measures the average slope across exactly that gap. If you let $h$ go toward 0, this turns into the slope at exactly this point. That slope is called the instantaneous rate of change and is the value of the derivative at that spot.

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

The derivative is the limit of the difference quotient as the gap

h

shrinks toward 0.

The slope as a new function

For $f(x) = x^2$ you can work the limit out yourself: $(x + h)^2 - x^2 = 2xh + h^2$ , and divided by $h$ that leaves $2x + h$ . As $h$ shrinks toward 0, only $2x$ remains. So the slope at position $x$ is always $2x$ , written $f'(x) = 2x$ . At $x = 3$ the slope is therefore $6$ , and at $x = 0$ it is $0$ , which is where the lowest point of the sits. The derivative is itself a function again.

Exercises

0 of 6 solved

Time to try it yourself. You can't break anything, every attempt counts.

What is the line called that touches the graph at one spot and has exactly its slope there?

What does the instantaneous rate of change at a point describe?

For $f(x) = x^2$ we have $f'(x) = 2x$ . What is the slope at $x = 4$ ?

Put the steps in the right order for how a secant becomes a tangent.

1Choose a fixed point on the graph
2The secant turns into the tangent
3Compute the slope of the secant for smaller and smaller gaps h
4Place a second point next to it and draw the secant
5Let the gap h go toward 0

For $f(x) = x^2$ with $f'(x) = 2x$ : at which position $x$ is the slope 0?

Match each value of the derivative to how the graph runs at that spot.

positive slope

negative slope

slope 0

Where this leads

Rules of differentiation Finding the tangent

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