Worked examples
The question
Question: find dy for y = x² when x = 3 and dx = 0.1.
y = x2
1 The rule
dx is a small change in x.
dy is the change the tangent line gives: the slope times dx.
dy = f′(x) dx
dy = f′(x) dx
2 Find f′(x)
Use the power rule on each term.
f′(x) = 2x^(2 − 1)
Now we have f′(x) = 2x.
= 2x
3 Put the numbers in
The slope at x = 3 is 6.
f′(3) = 2(3) = 6
So dy = 0.6.
dy = 6 × 0.1 = 0.6
Careful!
dy is the slope times dx.
Do not forget the times 0.1.
Careful: not dy = f′(3) = 6 ✗
4 Check
Let's check with the real change: f at 3.1, minus f at 3, is 0.61 on a calculator.
Close to 0.6.
Right!
Δy = f(3.1) − f(3) = 0.61
Let's look back.
First: the rule.
Then: find f′(x).
Then: put the numbers in.
Last: check.
Answer Careful: not dy = f′(3) = 6 ✗
▶ Watch it solved
The question
Question: find dy for y = x² when x = 3 and dx = 0.1, and compare it with Δy.
y = x2
1 The rule
dx is a small change in x.
dy is the change the tangent line gives: the slope times dx.
dy = f′(x) dx
dy = f′(x) dx
2 Find f′(x)
Use the power rule on each term.
f′(x) = 2x^(2 − 1)
Now we have f′(x) = 2x.
= 2x
3 Put the numbers in
The slope at x = 3 is 6.
f′(3) = 2(3) = 6
So dy = 0.6.
dy = 6 × 0.1 = 0.6
Careful!
dy is the slope times dx.
Do not forget the times 0.1.
Careful: not dy = f′(3) = 6 ✗
4 The real change
Δy is the real change: f at 3.1, minus f at 3.
Δy = f(3.1) − f(3)
A calculator gives f of 3.1: 9.61.
And f of 3 is 9.
= 9.61 − 9
So Δy = 0.61.
Very close to dy, 0.6.
= 0.61
5 Check
Let's check: dy and delta y differ by only 0.01.
A tiny gap, because dx is small.
Right!
dy − Δy = −0.01: tiny
Let's look back.
First: the rule.
Then: find f′(x).
Then: put the numbers in.
Then: the real change.
Last: check.
Answer Careful: not dy = f′(3) = 6 ✗
▶ Watch it solved
The question
Question: The circle has radius 8 cm, measured with a possible error of 0.1 cm.
Estimate the error in the area.
1 Draw and name
The radius is r = 8 cm.
r = 8 cm
The possible error in it is dr = 0.1 cm: a small change.
dr = 0.1 cm
2 Find dA
The area of a circle is A = π r².
A = πr2
Take the derivative, times dr: dA = 2πr dr.
dA = 2πr dr
The change in the area is about the derivative times the change in the radius.
dA = (derivative) × dr
3 Put the numbers in
Put in r = 8 and dr = 0.1.
dA = 2π(8)(0.1)
That is 1.6π, about 5.03 cm².
= 1.6π ≈ 5.03 cm²
Careful!
A small error in the radius makes a bigger error in the area.
Careful: dA is not dr: the error grows
4 Check
Let's check with the real change: the area at 8.1, minus the area at 8, is 5.06 cm².
Close to our 5.03.
Right!
A(8.1) − A(8) = 5.06 cm²
Let's look back.
First: draw and name.
Then: find dA.
Then: put the numbers in.
Last: check.
Answer Careful: dA is not dr: the error grows
▶ Watch it solved
