Worked examples
The question
Question: find the derivative of f(x) = (x² + 1)(3x − 2).
f(x) = (x2 + 1)(3x − 2)
1 Name u and v
Call the first bracket u: x² + 1.
u = x2 + 1
Call the second bracket v: 3x − 2.
v = 3x − 2
2 Find u′ and v′
The power rule on u gives u′ = 2x.
u′ = 2x
On v it gives v′ = 3.
v′ = 3
3 Use the rule
The product rule: the derivative of u times v is u′ v plus u v′.
(uv)′ = u′v + uv′
Put them in.
f′(x) = (2x)(3x − 2) + (x2 + 1)(3)
Careful!
Not u′ times v′!
Each part needs the other bracket.
Careful: not u′ × v′ = 6x ✗
4 Open the brackets
What is in front of a bracket multiplies every part inside.
= 6x2 − 4x + 3x2 + 3
Terms are the parts between the + and − signs.
Like terms are the same kind: same letters, same powers.
All the x terms go together.
Plain numbers go together.
So f′(x) = 9 x² − 4x + 3.
= 9x2 − 4x + 3
5 Check
Let's check another way: multiply the brackets out first.
f(x) is 3 x³ − 2 x² + 3x − 2.
f(x) = 3x3 − 2x2 + 3x − 2
The power rule on that gives 9 x² − 4x + 3.
The same!
Right!
f′(x) = 9x2 − 4x + 3
Let's look back.
First: name u and v.
Then: find u′ and v′.
Then: use the rule.
Then: open the brackets.
Last: check.
▶ Watch it solved
The question
Question: find the derivative of f(x) = (2 x² + x)(x − 3).
f(x) = (2x2 + x)(x − 3)
1 Name u and v
Call the first bracket u: 2 x² + x.
u = 2x2 + x
Call the second bracket v: x − 3.
v = x − 3
2 Find u′ and v′
The power rule on u gives u′ = 4x + 1.
u′ = 4x + 1
On v it gives v′ = 1.
v′ = 1
3 Use the rule
The product rule: the derivative of u times v is u′ v plus u v′.
(uv)′ = u′v + uv′
Put them in.
f′(x) = (4x + 1)(x − 3) + (2x2 + x)(1)
Careful!
Not u′ times v′!
Each part needs the other bracket.
Careful: not u′ × v′ = 4x + 1 ✗
4 Open the brackets
What is in front of a bracket multiplies every part inside.
= 4x2 − 12x + x − 3 + 2x2 + x
Terms are the parts between the + and − signs.
Like terms are the same kind: same letters, same powers.
All the x terms go together.
Plain numbers go together.
So f′(x) = 6 x² − 10x − 3.
= 6x2 − 10x − 3
5 Check
Let's check another way: multiply the brackets out first.
f(x) is 2 x³ − 5 x² − 3x.
f(x) = 2x3 − 5x2 − 3x
The power rule on that gives 6 x² − 10x − 3.
The same!
Right!
f′(x) = 6x2 − 10x − 3
Let's look back.
First: name u and v.
Then: find u′ and v′.
Then: use the rule.
Then: open the brackets.
Last: check.
▶ Watch it solved
The question
Question: find the derivative of f(x) = (x² + 1)/(x − 2).
f(x) = x2 + 1x − 2
1 Name u and v
Call the top u: x² + 1.
u = x2 + 1
Call the bottom v: x − 2.
v = x − 2
2 Find u′ and v′
The power rule on u gives u′ = 2x.
u′ = 2x
On v it gives v′ = 1.
v′ = 1
3 Use the rule
The quotient rule: u′ v minus u v′, all over v squared.
The order on top matters: u′ v comes first.
(uv)′ = u′v − uv′v2
Careful!
Not u v′ minus u′ v!
Start with u′: the top's derivative comes first.
Careful: not uv′ − u′v ✗
Put them in.
The bottom stays as v squared.
f′(x) = (2x)(x − 2) − (x2 + 1)(1)(x − 2)2
4 Tidy the top
What is in front of a bracket multiplies every part inside.
= 2x2 − 4x − (x2 + 1)(x − 2)2
A minus in front of a bracket changes the sign of every part inside.
= 2x2 − 4x − x2 − 1(x − 2)2
Terms are the parts between the + and − signs.
Like terms are the same kind: same letters, same powers.
All the x terms go together.
Plain numbers go together.
So f′(x) = (x² − 4x − 1)/(x − 2)².
= x2 − 4x − 1(x − 2)2
5 Check
Let's check: the derivative is the slope.
At x = 0, the rule gives −0.25.
f′(0) = −1(−2)2 = −14
Now the slope between two very close points: x = 0 and x = 0.001.
A calculator gives about −0.25.
Very close to −0.25.
Right!
f(0.001) − f(0)0.001 ≈ −0.25
Let's look back.
First: name u and v.
Then: find u′ and v′.
Then: use the rule.
Then: tidy the top.
Last: check.
Answer = x2 − 4x − 1(x − 2)2
▶ Watch it solved
