Derivative of y = (x + 3)^2 (x + 1) using logarithms Step by Step

Step by step

The question

Question: find dy/dx for y = (x + 3)²(x + 1), using logarithms.

y = (x + 3)²(x + 1)

The log rules turn a product into a sum, a fraction into a difference, and bring powers down.

ln(ab) = ln a + ln b

lnab = ln a − ln b

ln(a^n) = n ln a

Careful!

The ln of a product is a sum of lns, not a product.

Careful: not ln(ab) = ln a × ln b ✗

1 Take ln of both sides

Take ln of both sides.

ln y = ln((x + 3)²(x + 1))

2 Use the log rules

Use the log rules: each power comes down, multiplying becomes adding.

ln y = 2 ln(x + 3) + ln(x + 1)

3 The derivative of each side

The left side: 1/y, times dy/dx, by the chain rule.

Each ln(x + c) gives 1/x + c.

(ln y)′ = 1y × dydx

(ln(x + c))′ = 1x + c

Each power waits in front.

1ydydx = 2x + 3 + 1x + 1

4 Multiply by y

Multiply both sides by y.

dydx = y(2x + 3 + 1x + 1)

5 Put y back

Put y back.

dy/dx = (x + 3)² (x + 1)(2/(x + 3) + 1/(x + 1)).

Books leave it like this.

dydx = (x + 3)²(x + 1)(2x + 3 + 1x + 1)

6 Check

Let's check: the derivative is the slope.

At x = 2, our answer gives 55, with a calculator.

dy/dx at x = 2: 55

Now the slope between two very close points: x = 2 and x = 2.001.

A calculator gives about 55.01.

Very close to 55.

Right!

y(2.001) − y(2)0.001 ≈ 55.01

Let's look back.

First: take ln of both sides.

Then: use the log rules.

Then: the derivative of each side.

Then: multiply by y.

Then: put y back.

Last: check.

Answer