For y = x^2 + 3x, what is the differential dy when x = 1 and dx =… Step by Step

Step by step

The question

Question: find dy for y = x² + 3x when x = 1 and dx = 0.01.

y = x² + 3x

A small change in x makes a small change in y.

There are four choices.

First we solve it, the normal way.

Then we pick.

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

2 Find f′(x)

Use the power rule on each term.

f′(x) = 2x^(2 − 1) + 3 × 1

Now we have f′(x) = 2x + 3.

= 2x + 3

3 Put the numbers in

The slope at x = 1 is 5.

f′(1) = 2(1) + 3 = 2 + 3 = 5

So dy = 0.05.

dy = 5 × 0.01 = 0.05

Careful!

dy is the slope times dx.

Do not forget the times 0.01.

Careful: not dy = f′(1) = 5 ✗

4 Check

Let's check with the real change: f at 1.01, minus f at 1, is 0.0501 on a calculator.

Close to 0.05.

Right!

Δy = f(1.01) − f(1) = 0.0501

5 Pick the choice

Our answer is dy = 0.05.

That is choice C.

C) 0.05

6 Why not the others

A is 0.0501.

Wrong: that is Δy, the exact change: dy uses the slope.

A) 0.0501 ✗ that is Δy, the exact change: dy uses the slope

B is 5.

Wrong: that is dy/dx, the slope.

That is the Careful mistake from before.

B) 5 ✗ that is dy/dx, the slope

D is 0.02.

Wrong: lost the 3x: 2x × dx.

D) 0.02 ✗ lost the 3x: 2x × dx

Let's look back.

First: the rule.

Then: find f′(x).

Then: put the numbers in.

Then: check.

Then: pick the choice.

Last: why not the others.

Answer