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Derivative of 4log_3(2x + 1)

Step by step, with the reason for every step · Derivatives of bˣ / Derivatives of logarithms · Calculus 1

Step by step

The question

Question: find the derivative of f(x) = 4log₃(2x + 1).

f(x) = 4log3(2x + 1)

1 Write it with e

Change of base: log base b of x is ln x over ln b.

logb(x) = ln xln b

So it is ln of 2x + 1, over ln 3.

ln 3 is just a number.

log3(2x + 1) = ln(2x + 1)ln 3

2 The rule

The number 1 over ln 3 waits.

ln u gives 1/u, times the inside's slope, 2.

ddxln(2x + 1)ln 3 = 1ln 3 × 1(2x + 1) × 2

Careful!

That is only true for ln.

For log base 3, divide by ln 3 too.

Careful: not log3(x) → 1x ✗

3 Tidy up

Multiply into one fraction, the 4 on top: f′(x) = 8 over ((2x + 1) ln 3).

f′(x) = 8(2x + 1) ln 3

4 Check

Let's check: the derivative is the slope.

At x = 1, our answer gives 2.4273, with a calculator.

f′(1) = 2.4273

Now the slope between two very close points: x = 1 and x = 1.001.

A calculator gives about 2.43.

Very close to 2.4273.

Right!

f(1.001) − f(1)0.001 ≈ 2.43

Let's look back.

First: write it with e.

Then: the rule.

Then: tidy up.

Last: check.

Answer
f′(x) = 8(2x + 1) ln 3

Your turn

Try this one yourself, then check:

Find the derivative of f(x) = 5log3(2x + 3)
Show the answer
f′(x) = 10(2x + 3) ln 3

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