10 min readPathway sheetOften a 3-mark question; sometimes a 5-mark question with two power functions added

Logarithmic differentiation

For y = u^v (a variable raised to a variable power) or a long product, take log of both sides first, then differentiate.

When to use itFor [u(x)]^(v(x)), such as xˣ or (sin x)^(cos x), and for long products or quotients

Key identitylog(u^v) = v log u

ResultIf y = u^v, then dy/dx = u^v [v′ log u + v u′/u]

Sum of two powersFor y = u + v, differentiate u and v separately and add. Never take the log of a sum

The method

Here u^v means u raised to the power v.

1
Take log of both sides: for y = u^v, log y = v log u.
2
Differentiate both sides with respect to x: (1/y) dy/dx = v′ log u + v × u′/u.
3★
Multiply both sides by y.
4
Replace y by the original expression u^v.

★ key step

Example 1: a variable power

Problem

Differentiate y = x^(√x), x > 0.

Working

Take log: log y = √x log x
Differentiate (product rule): (1/y) dy/dx = [1/(2√x)] log x + √x × (1/x)
= (log x)/(2√x) + 1/√x = (log x + 2)/(2√x)
dy/dx = y × (2 + log x)/(2√x)

Answer

dy/dx = x^(√x) (2 + log x)/(2√x)

Example 2: a sum of two powers

Problem

Find dy/dx if y = xˣ + (sin x)ˣ, 0 < x < π.

Working

Let u = xˣ and v = (sin x)ˣ. Then dy/dx = du/dx + dv/dx.
log u = x log x, so (1/u) du/dx = log x + 1, and du/dx = xˣ(1 + log x).
log v = x log(sin x), so (1/v) dv/dx = log(sin x) + x × (cos x/sin x).
dv/dx = (sin x)ˣ [log(sin x) + x cot x]
Add the two parts.

Answer

dy/dx = xˣ(1 + log x) + (sin x)ˣ [log(sin x) + x cot x]

Example 3: a long product and quotient

Problem

Find dy/dx if y = (x + 1)²(x − 2)³/(x + 3)⁴, x > 2.

Working

Take log: log y = 2 log(x + 1) + 3 log(x − 2) − 4 log(x + 3)
Differentiate: (1/y) dy/dx = 2/(x + 1) + 3/(x − 2) − 4/(x + 3)
Multiply by y.

Answer

dy/dx = [(x + 1)²(x − 2)³/(x + 3)⁴] × [2/(x + 1) + 3/(x − 2) − 4/(x + 3)]

Log of a product vs log of a sum

Log of a productlog(uv) = log u + log v. Allowed: this is why logs make products easy.

Log of a sumlog(u + v) is not log u + log v. For y = u + v, find du/dx and dv/dx separately.

This one mistake costs the most marks in this topic.

In real life

Before calculators, engineers and scientists multiplied large numbers with log tables, turning products into sums. Logarithmic differentiation uses the same trick to tame hard powers and products.

Exam angle

2 marks: “Find dy/dx if y = xˣ, x > 0.” log y = x log x. (1/y) dy/dx = log x + 1. So dy/dx = xˣ(1 + log x).

3 marks: “Find dy/dx if y = (sin x)^(cos x), 0 < x < π/2.” log y = cos x log(sin x). (1/y) dy/dx = −sin x log(sin x) + cos x cot x. So dy/dx = (sin x)^(cos x) [cos x cot x − sin x log(sin x)].

5 marks: “Find dy/dx if y = x^(cos x) + (cos x)^(sin x), 0 < x < π/2.” u = x^(cos x): du/dx = x^(cos x)[(cos x)/x − sin x log x]. v = (cos x)^(sin x): dv/dx = (cos x)^(sin x)[cos x log(cos x) − sin x tan x]. dy/dx = du/dx + dv/dx.

Remember it

Power on power? Take log first. Sum of powers? Split first.