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Find dy for y = x^3 - 2x when x = 2 and dx = 0.1, and compare with Δy

Step by step, with the reason for every step · Differentials · Calculus 1

Step by step

The question

Question: find dy for y = x³ − 2x when x = 2 and dx = 0.1, and compare it with Δy.

y = x3 − 2x

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) − 2 × 1

Now we have f′(x) = 3x² − 2.

= 3x2 − 2

3 Put the numbers in

The slope at x = 2 is 10.

f′(2) = 3(2)2 − 2 = 12 − 2 = 10

So dy = 1.

dy = 10 × 0.1 = 1

Careful!

dy is the slope times dx.

Do not forget the times 0.1.

Careful: not dy = f′(2) = 10 ✗

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: 5.061.

And f of 2 is 4.

= 5.061 − 4

So Δy = 1.061.

Very close to dy, 1.

= 1.061

5 Check

Let's check: dy and delta y differ by only 0.061.

A tiny gap, because dx is small.

Right!

dy − Δy = −0.061: 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) = 10 ✗

Your turn

Try this one yourself, then check:

Find dy for y = 3x3 − x
Show the answer
dy = 16, Δy = 17.104

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