Step by step
The question
Question: find dy/dx for y = log₂(x² + 3).
y = log2(x2 + 3)
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 x² + 3, over ln 2.
ln 2 is just a number.
log2(x2 + 3) = ln(x2 + 3)ln 2
2 The rule
The number 1 over ln 2 waits.
ln u gives 1/u, times the inside's slope, 2x.
ddxln(x2 + 3)ln 2 = 1ln 2 × 1(x2 + 3) × 2x
Careful!
That is only true for ln.
For log base 2, divide by ln 2 too.
Careful: not log2(x) → 1x ✗
3 Tidy up
Multiply into one fraction: dy/dx = 2x over ((x² + 3) ln 2).
dydx = 2x(x2 + 3) ln 2
4 Check
Let's check: the derivative is the slope.
At x = 1, our answer gives 0.7213, with a calculator.
dy/dx at x = 1: 0.7213
Now the slope between two very close points: x = 1 and x = 1.001.
A calculator gives about 0.72.
Very close to 0.7213.
Right!
y(1.001) − y(1)0.001 ≈ 0.72
Let's look back.
First: write it with e.
Then: the rule.
Then: tidy up.
Last: check.
Answer
dydx = 2x(x2 + 3) ln 2