Step by step
The question
Question: find dy for y = 3x² + 2 when x = 2 and dx = 0.1, and compare it with Δy.
y = 3x2 + 2
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) = 3 × 2x^(2 − 1) + 0
Now we have f′(x) = 6x.
= 6x
3 Put the numbers in
The slope at x = 2 is 12.
f′(2) = 6(2) = 12
So dy = 1.2.
dy = 12 × 0.1 = 1.2
Careful!
dy is the slope times dx.
Do not forget the times 0.1.
Careful: not dy = f′(2) = 12 ✗
4 The real change
Δy is the real change: f at 2.1, minus f at 2.
Δy = f(2.1) − f(2)
A calculator gives f of 2.1: 15.23.
And f of 2 is 14.
= 15.23 − 14
So Δy = 1.23.
Very close to dy, 1.2.
= 1.23
5 Check
Let's check: dy and delta y differ by only 0.03.
A tiny gap, because dx is small.
Right!
dy − Δy = −0.03: 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′(2) = 12 ✗