Step by step
The question
Question: find dy for y = x³ when x = 1 and dx = 0.05, and compare it with Δy.
y = x3
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) = 3x^(3 − 1)
Now we have f′(x) = 3x².
= 3x2
3 Put the numbers in
The slope at x = 1 is 3.
f′(1) = 3(1)2 = 3
So dy = 0.15.
dy = 3 × 0.05 = 0.15
Careful!
dy is the slope times dx.
Do not forget the times 0.05.
Careful: not dy = f′(1) = 3 ✗
4 The real change
Δy is the real change: f at 1.05, minus f at 1.
Δy = f(1.05) − f(1)
A calculator gives f of 1.05: 1.15762.
And f of 1 is 1.
= 1.15762 − 1
So Δy = 0.15762.
Very close to dy, 0.15.
= 0.15762
5 Check
Let's check: dy and delta y differ by only 0.00762.
A tiny gap, because dx is small.
Right!
dy − Δy = −0.00762: 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′(1) = 3 ✗