Worked examples
The question
Question: find dy/dx for y = (3x + 1)⁴.
y = (3x + 1)4
1 The inside
There is something inside: 3x + 1.
Call it u.
u = 3x + 1
Then y is just u⁴.
y = u4
2 The outside
First the outside.
Use the power rule on u⁴.
dydu = 4u^(4 − 1)
We get 4 u³.
= 4u3
3 The inside's slope
Now the inside.
The power rule on 3x + 1 gives 3.
dudx = 3
4 Multiply
The chain rule: multiply the two.
dy/du times du/dx.
dydx = dydu × dudx
Put the inside back in place of u.
= 4(3x + 1)3 × 3
Careful!
Do not stop here!
We must multiply by the inside's slope, 3.
Careful: not 4(3x + 1)3 ✗
Multiply the numbers in front.
So dy/dx = 12 (3x + 1)³.
= 12(3x + 1)3
4 × 3 = 12
5 Check
Let's check: the derivative is the slope.
At x = 0, the rule gives 12.
dy/dx at x = 0: 12
Now the slope between two very close points: x = 0 and x = 0.001.
A calculator gives about 12.05.
Very close to 12.
Right!
y(0.001) − y(0)0.001 ≈ 12.05
Let's look back.
First: the inside.
Then: the outside.
Then: the inside's slope.
Then: multiply.
Last: check.
▶ Watch it solved
The question
Question: find dy/dx for y = (x² − 5)³.
y = (x2 − 5)3
1 The inside
There is something inside: x² − 5.
Call it u.
u = x2 − 5
Then y is just u³.
y = u3
2 The outside
First the outside.
Use the power rule on u³.
dydu = 3u^(3 − 1)
We get 3 u².
= 3u2
3 The inside's slope
Now the inside.
The power rule on x² − 5 gives 2x.
dudx = 2x
4 Multiply
The chain rule: multiply the two.
dy/du times du/dx.
dydx = dydu × dudx
Put the inside back in place of u.
= 3(x2 − 5)2 × 2x
Careful!
Do not stop here!
We must multiply by the inside's slope, 2x.
Careful: not 3(x2 − 5)2 ✗
Multiply the numbers in front.
So dy/dx = 6x (x² − 5)².
= 6x(x2 − 5)2
3 × 2x = 6x
5 Check
Let's check: the derivative is the slope.
At x = undefined, the rule gives NaN.
dy/dx at x = undefined: NaN
Now the slope between two very close points: x = undefined and x = NaN.
A calculator gives about NaN.
Very close to NaN.
Right!
y(NaN) − y(undefined)0.001 ≈ NaN
Let's look back.
First: the inside.
Then: the outside.
Then: the inside's slope.
Then: multiply.
Last: check.
▶ Watch it solved
The question
Question: find dy/dx for y = 1/(x² + 1)².
y = 1(x2 + 1)2
1 The inside
There is something inside: x² + 1.
Call it u.
u = x2 + 1
Then y is just u⁻².
Written as a power of u.
y = u−2
u to a power at the bottom is a minus power on top.
1u2 = u−2
2 The outside
First the outside.
Use the power rule on u⁻².
dydu = (−2)u^(−2 − 1)
We get −2 u⁻³.
= −2u−3
Take 1 away from the power: −2 − 1 = −3.
−2 − 1 = −3
3 The inside's slope
Now the inside.
The power rule on x² + 1 gives 2x.
dudx = 2x
4 Multiply
The chain rule: multiply the two.
dy/du times du/dx.
dydx = dydu × dudx
Put the inside back in place of u.
= −2(x2 + 1)−3 × 2x
Careful!
Do not stop here!
We must multiply by the inside's slope, 2x.
Careful: not −2(x2 + 1)−3 ✗
Multiply the numbers in front.
Then write the minus power at the bottom.
So dy/dx = −4x/(x² + 1)³.
= −4x(x2 + 1)3
−2 × 2x = −4x
5 Check
Let's check: the derivative is the slope.
At x = 0, the rule gives 0.
dy/dx at x = 0: 0
Now the slope between two very close points: x = 0 and x = 0.001.
A calculator gives about 0.
Very close to 0.
Right!
y(0.001) − y(0)0.001 ≈ 0
Let's look back.
First: the inside.
Then: the outside.
Then: the inside's slope.
Then: multiply.
Last: check.
▶ Watch it solved
