Worked examples
The question
Question: find f′(x) for f(x) = x², from the definition.
f(x) = x2
1 The idea
Look at the curve y = x².
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.
f′(x) = lim h → 0 f(x + h) − f(x)h
2 Find f(x + h)
Put x + h in place of x, in brackets.
f(x + h) = (x + h)2
(x + h)² means (x + h) times (x + h).
Open it: x² + 2xh + h².
(x + h)2 = x2 + 2xh + h2
Careful!
The 2xh in the middle must not be lost.
Careful: not (x + h)2 = x² + h² ✗
Now we have x² + 2xh + h².
= x2 + 2xh + h2
3 Take away f(x)
Now take away f(x).
It goes in brackets: − (x²).
f(x + h) − f(x) = x2 + 2xh + h2 − (x2)
A minus in front of a bracket changes the sign of every part inside.
= x2 − x2
Every term of f(x) meets its opposite.
They make zero: cross them out.
Write what is left.
Every term left has an h.
= 2xh + h2
4 Divide by h
Divide by the run, h.
Every term on top has an h, so take it out: 2xh + h² is h(2x + h).
f(x + h) − f(x)h = h(2x + h)h
h is close to 0, but not 0.
So h/h is 1.
Cross them out.
Write what is left.
= 2x + h
5 Let h go to 0
Now let h go to 0.
f′(x) = lim h → 0 (2x + h)
Every term with an h becomes 0.
Cross them out.
So the derivative is f′(x) = 2x.
= 2x
6 Check
Let's check on the picture.
P is at x = 1.
The rule gives f′(1) = 2.
f′(1) = 2(1) = 2
From P, go right 1.
Then up 2.
We land on the 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.
▶ Watch it solved
The question
Question: find f′(x) for f(x) = x² − 3x, from the definition.
f(x) = x2 − 3x
1 The idea
Look at the curve y = x² − 3x.
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.
f′(x) = lim h → 0 f(x + h) − f(x)h
2 Find f(x + h)
Put x + h in place of x, in brackets.
f(x + h) = (x + h)2 − 3(x + h)
(x + h)² means (x + h) times (x + h).
Open it: x² + 2xh + h².
(x + h)2 = x2 + 2xh + h2
Careful!
The 2xh in the middle must not be lost.
Careful: not (x + h)2 = x² + h² ✗
What is in front of a bracket multiplies every part inside.
Now we have x² + 2xh + h² − 3x − 3h.
= x2 + 2xh + h2 − 3x − 3h
3 Take away f(x)
Now take away f(x).
It goes in brackets: − (x² − 3x).
f(x + h) − f(x) = x2 + 2xh + h2 − 3x − 3h − (x2 − 3x)
A minus in front of a bracket changes the sign of every part inside.
= x2 − 3x − 3h − x2 + 3x
Every term of f(x) meets its opposite.
They make zero: cross them out.
Write what is left.
Every term left has an h.
= 2xh + h2 − 3h
4 Divide by h
Divide by the run, h.
Every term on top has an h, so take it out: 2xh + h² − 3h is h(2x + h − 3).
f(x + h) − f(x)h = h(2x + h − 3)h
h is close to 0, but not 0.
So h/h is 1.
Cross them out.
Write what is left.
= 2x + h − 3
5 Let h go to 0
Now let h go to 0.
f′(x) = lim h → 0 (2x + h − 3)
Every term with an h becomes 0.
Cross them out.
So the derivative is f′(x) = 2x − 3.
= 2x − 3
6 Check
Let's check on the picture.
P is at x = 1.
The rule gives f′(1) = −1.
f′(1) = 2(1) − 3 = 2 − 3 = −1
From P, go right 1.
Then down 1.
We land on the 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.
▶ Watch it solved
The question
Question: find f′(x) for f(x) = 2x² + x, from the definition, and the slope at x = 1.
f(x) = 2x2 + x
1 The idea
Look at the curve y = 2x² + x.
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.
f′(x) = lim h → 0 f(x + h) − f(x)h
2 Find f(x + h)
Put x + h in place of x, in brackets.
f(x + h) = 2(x + h)2 + (x + h)
(x + h)² means (x + h) times (x + h).
Open it: x² + 2xh + h².
(x + h)2 = x2 + 2xh + h2
Careful!
The 2xh in the middle must not be lost.
Careful: not (x + h)2 = x² + h² ✗
Open the power first.
= 2(x2 + 2xh + h2) + (x + h)
What is in front of a bracket multiplies every part inside.
Now we have 2x² + 4xh + 2h² + x + h.
= 2x2 + 4xh + 2h2 + x + h
3 Take away f(x)
Now take away f(x).
It goes in brackets: − (2x² + x).
f(x + h) − f(x) = 2x2 + 4xh + 2h2 + x + h − (2x2 + x)
A minus in front of a bracket changes the sign of every part inside.
= 2x2 + 4xh + 2h2 + x + h − 2x2 − x
Every term of f(x) meets its opposite.
They make zero: cross them out.
Write what is left.
Every term left has an h.
= 4xh + 2h2 + h
4 Divide by h
Divide by the run, h.
Every term on top has an h, so take it out: 4xh + 2h² + h is h(4x + 2h + 1).
f(x + h) − f(x)h = h(4x + 2h + 1)h
h is close to 0, but not 0.
So h/h is 1.
Cross them out.
Write what is left.
= 4x + 2h + 1
5 Let h go to 0
Now let h go to 0.
f′(x) = lim h → 0 (4x + 2h + 1)
Every term with an h becomes 0.
Cross them out.
So the derivative is f′(x) = 4x + 1.
= 4x + 1
6 The slope at x = 1
Put 1 in.
The slope at x = 1 is 5.
f′(1) = 4(1) + 1 = 4 + 1 = 5
7 Check
From P, go right 1.
Then up 5.
We land on the 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.
Then: the slope at x = 1.
Last: check.
▶ Watch it solved
