Two basic results cover this lesson: d/dx of ex is ex, and d/dx of ln x is 1/x. For anything more than a plain x inside, add the chain rule: d/dx of ef(x) is f′(x)ef(x), and d/dx of ln f(x) is f′(x)/f(x).
Confirm on the current 0606 syllabus page which logarithmic and exponential functions you are expected to differentiate. This lesson assumes ex and ln x, which also appear in exponential and logarithmic reasoning.
How do the forms work in practice?
| y | dy/dx |
|---|---|
| ex | ex |
| ekx | kekx |
| ef(x) | f′(x)ef(x) |
| ln x | 1/x |
| ln(ax + b) | a/(ax + b) |
| ln f(x) | f′(x)/f(x) |
The exponential keeps its own form, and the inside supplies a multiplier. The logarithm turns into a fraction with the inside’s derivative on top and the inside itself underneath.
Worked example
Differentiate (a) y = e2x + ln(3x − 1), and (b) y = x e−x.
(a) For e2x, the inside is 2x, with derivative 2. So the derivative is 2e2x.
For ln(3x − 1), the inside is 3x − 1, with derivative 3. So the derivative is 3/(3x − 1).
dy/dx = 2e2x + 3/(3x − 1).
(b) This is a product. Let u = x and v = e−x. Then u′ = 1 and v′ = −e−x.
dy/dx = u′v + uv′ = e−x + x(−e−x) = e−x − xe−x = (1 − x)e−x.
Quick sense check: at x = 0 the gradient is 1, and y = xe−x is close to y = x near the origin, so this fits.
The mistake to watch for
The tempting shortcut is to apply the “basic” result to the whole expression and ignore the inside.
Mistaken working: y = ln(x² + 4) → dy/dx = 1/(x² + 4)
The student wrote “1 over the inside” and forgot the inside’s derivative.
The numerator must be the derivative of x² + 4, which is 2x, so the correct answer is 2x/(x² + 4).
A related slip is to write e2x → 2x e2x−1. The power rule does not apply to ex, because x is in the power, not the base. The correct derivative is 2e2x.
Check yourself
1. Differentiate y = e5x.
Show answer
The inside is 5x with derivative 5, so dy/dx = 5e5x.
2. Differentiate y = ln(4x + 1) and find the gradient at x = 0.
Show answer
dy/dx = 4/(4x + 1). At x = 0: 4/1 = 4.
3. Differentiate y = x²ex, factorise, and find the gradient at x = 1, leaving e in the answer.
Show answer
Let u = x² and v = ex, so u′ = 2x and v′ = ex.
dy/dx = 2xex + x²ex = x(x + 2)ex.
At x = 1: 1 × 3 × e = 3e (about 8.15).
Where this leads next
The final technique lesson, checking a derivative by comparing local gradients, is a good test for answers involving ex and ln. The calculus shape and rate explorer draws the curve and its gradient, and the non-calculator working trainer helps with leaving answers in terms of e.
Students often meet these functions late and feel behind on them. A teacher in online one-to-one Additional Mathematics tuition can start from the rules you already know and add the new ones at your pace.