How do we differentiate exponential and logarithmic functions?
The derivative of e^x is itself, the derivative of ln x is 1/x, and the chain rule extends both to composite functions.
The derivatives of e^x and ln x, the chain-rule extensions for e^{f(x)} and ln(f(x)), and worked applications including growth models and stationary points.
Reviewed by: AI editorial process; not yet individually human-reviewed
Have a quick question? Jump to the Q&A page
Jump to a section
What this dot point is asking
The exponential function is special: it is its own derivative. The natural logarithm differentiates to . These two results, combined with the chain rule, let you differentiate every exponential and logarithmic model in Stage 2.
The base results
Chain-rule extensions
When the exponent or the argument is itself a function of , multiply by its derivative:
Combining with the product and quotient rules
These derivatives frequently sit inside products and quotients.
Using the log laws to simplify first
Before differentiating a complicated log, expand it with the log laws - this is far easier than the quotient or product rule.
For , rewrite as , then
Common errors
Why it matters
These derivatives are the basis of every exponential growth-and-decay and logarithmic application in Stage 2, and they connect directly to the antidifferentiation results in Topic 3 (since and are simply these rules reversed).
Exam-style practice questions
Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2017 SACE Stage 22 marksFind dy/dx for y = e^(x^2). There is no need to simplify your answer.Show worked answer →
This is the exponential chain rule d/dx e^(g(x)) = g'(x) e^(g(x)). The inner function is g(x) = x^2 with g'(x) = 2x.
So dy/dx = 2x e^(x^2).
Marks: one for recognising the derivative of e^(g(x)) keeps the same exponential factor e^(x^2), and one for multiplying by the inner derivative 2x. A frequent error is writing e^(x^2) alone and omitting the factor 2x that the chain rule supplies.
2017 SACE Stage 22 marksFind dy/dx if y = x e^x.Show worked answer →
This is a product of u = x and v = e^x, so use the product rule (uv)' = u'v + uv'.
Here u' = 1 and v' = e^x (the exponential is its own derivative).
dy/dx = (1)(e^x) + (x)(e^x) = e^x + x e^x = e^x (1 + x).
Marks: one for the correct product-rule set-up, one for the simplified derivative e^x(1 + x). Factoring out e^x is tidy and is exactly the form needed if the next part asks for stationary points or an antiderivative of x e^x.