Worked examples
The question
Question: find the derivative of f(x) = x⁵.
f(x) = x5
1 The rule
From the definition we found: x² gives 2x, and x³ gives 3x².
x2 → 2x
x3 → 3x2
See the pattern?
The power comes down in front, and the power goes down by 1.
This is the power rule.
ddxxn = nxn − 1
We write the derivative as f′(x), "f prime of x".
f′(x): the derivative of f(x)
2 Each term
Use the rule on each term.
f′(x) = 5x^(5 − 1)
Careful!
The power must go down by 1 too.
Careful: not x5 → 5x5 ✗
Work out each one.
So f′(x) = 5 x⁴.
= 5x4
3 Check
Let's check: the derivative is the slope.
At x = 1, the rule gives 5.
f′(1) = 5
Now the slope between two very close points: x = 1 and x = 1.001.
A calculator gives about 5.01.
Very close to 5.
Right!
f(1.001) − f(1)0.001 ≈ 5.01
Let's look back.
First: the rule.
Then: each term.
Last: check.
▶ Watch it solved
The question
Question: find the derivative of f(x) = 5 x³ − 2 x² + x − 8.
f(x) = 5x3 − 2x2 + x − 8
1 The rule
From the definition we found: x² gives 2x, and x³ gives 3x².
x2 → 2x
x3 → 3x2
See the pattern?
The power comes down in front, and the power goes down by 1.
This is the power rule.
ddxxn = nxn − 1
We write the derivative as f′(x), "f prime of x".
f′(x): the derivative of f(x)
2 Each term
Use the rule on each term.
A number in front just waits, and multiplies.
f′(x) = 5 × 3x^(3 − 1) − 2 × 2x^(2 − 1) + 1 + 0
A number alone gives 0.
Its graph is a flat line, and a flat line has slope 0.
a number alone → 0
Careful!
A number alone does not stay.
It gives 0.
Careful: not −8 → −8 ✗
Work out each one.
So f′(x) = 15 x² − 4x + 1.
= 15x2 − 4x + 1
3 Check
Let's check: the derivative is the slope.
At x = 1, the rule gives 12.
f′(1) = 15 − 4 + 1 = 12
Now the slope between two very close points: x = 1 and x = 1.001.
A calculator gives about 12.01.
Very close to 12.
Right!
f(1.001) − f(1)0.001 ≈ 12.01
Let's look back.
First: the rule.
Then: each term.
Last: check.
▶ Watch it solved
The question
Question: find the derivative of f(x) = 6 x^(2/3).
f(x) = 6x^(2/3)
1 The rule
From the definition we found: x² gives 2x, and x³ gives 3x².
x2 → 2x
x3 → 3x2
See the pattern?
The power comes down in front, and the power goes down by 1.
This is the power rule.
ddxxn = nxn − 1
We write the derivative as f′(x), "f prime of x".
f′(x): the derivative of f(x)
2 Each term
Use the rule on each term.
A number in front just waits, and multiplies.
f′(x) = 6 × 23x^(2/3 − 1)
Take 1 away from the power: 2/3 − 1 = −1/3.
23 − 1 = −13
Careful!
The power must go down by 1 too.
Careful: not 6x^(2/3) → 4x^(2/3) ✗
Work out each one.
= 4x^(−1/3)
3 Tidy up
Books write it without minus or fraction powers.
So f′(x) = 4 over (x^(1/3)).
= 4x^(1/3)
4 Check
Let's check: the derivative is the slope.
At x = 1, the rule gives 4.
f′(1) = 4
Now the slope between two very close points: x = 1 and x = 1.001.
A calculator gives about 4.
Very close to 4.
Right!
f(1.001) − f(1)0.001 ≈ 4
Let's look back.
First: the rule.
Then: each term.
Then: tidy up.
Last: check.
▶ Watch it solved
