Step by step
The question
Question: find f′(x) for f(x) = 4x² − 8x + 1, from the definition.
1 The idea
Look at the curve y = 4x² − 8x + 1.
Take a point P on it.
How steep is the curve at P?
Take a second point Q, a little to the right: x + h.
Here h = 1.
The line through P and Q has slope rise over run.
The run is h.
Move Q closer: h = ½.
The line turns.
It gets closer to the curve's own steepness.
When h goes to 0, the line just touches the curve at P.
It is called the tangent.
Its slope is the derivative.
Written as a rule: the rise is f(x + h) minus f(x).
The run is h.
Then h goes to 0.
2 Find f(x + h)
Put x + h in place of x, in brackets.
(x + h)² means (x + h) times (x + h).
Open it: x² + 2xh + h².
Careful!
The 2xh in the middle must not be lost.
Open the power first.
What is in front of a bracket multiplies every part inside.
Now we have 4x² + 8xh + 4h² − 8x − 8h + 1.
3 Take away f(x)
Now take away f(x).
It goes in brackets: − (4x² − 8x + 1).
A minus in front of a bracket changes the sign of every part inside.
Every term of f(x) meets its opposite.
They make zero: cross them out.
Write what is left.
Every term left has an h.
4 Divide by h
Divide by the run, h.
Every term on top has an h, so take it out: 8xh + 4h² − 8h is h(8x + 4h − 8).
h is close to 0, but not 0.
So h/h is 1.
Cross them out.
Write what is left.
5 Let h go to 0
Now let h go to 0.
Every term with an h becomes 0.
Cross them out.
So the derivative is f′(x) = 8x − 8.
6 Check
Let's check on the picture.
P is at x = 1.
The rule gives f′(1) = 0.
From P, go right 1.
The slope is 0: no going up.
We stay on the flat tangent.
Right!
Let's look back.
First: the idea.
Then: find f(x + h).
Then: take away f(x).
Then: divide by h.
Then: let h go to 0.
Last: check.