Derivatives of exponential and logarithmic functions sit at the centre of the AP Calculus AB and BC syllabus, and the topic carries disproportionate weight relative to the time it takes to learn. In roughly one focused sitting, a candidate can internalise the rule set for differentiating ex, ax, ln x, and loga x, then move into differentiation rules that combine those building blocks with the chain, product, and quotient rules. The difficulty is not memorisation; it is recognition. AP exam writers like to bury a derivative of e2x or ln(cos x) inside a larger problem where the exponential or logarithmic part is a single layer among several. Candidates who treat the topic as a standalone unit often recognise it cold in isolation, then miss it inside a related-rates word problem. The five rules below, the worked examples, and the tactical notes are designed to fix that gap before it costs marks on the multiple-choice section or the free-response portion of either the AB or the BC exam.
The core derivative rules, written so they actually stick
Most AP Calculus textbooks list the derivative rules for exponential and logarithmic functions in a single boxed summary. Memorising the box is the easy part; the harder task is writing each rule in a form that you can re-derive on a blank piece of paper when you are three hours into the exam and your working memory is depleted. The five rules every AB and BC candidate should be able to reproduce from memory are listed below, and each is paired with the form in which it most often appears inside a larger expression.
- d/dx (ex) = ex. The function is its own derivative. This is the only elementary function with that property, and AP questions use it to test whether you notice that no chain-rule factor is needed when the exponent is exactly x.
- d/dx (ax) = ax · ln a. For any positive base a ≠ 1. The constant ln a sits out front; it is easy to forget and easy to misread as a coefficient of x.
- d/dx (ln x) = 1/x. Defined for x > 0. On the AP exam the domain restriction almost never appears as a graded point, but it shows up in curve-sketching free response where the domain is part of the justification.
- d/dx (loga x) = 1 / (x · ln a). The natural log of the base sits in the denominator, multiplied by x. Many candidates write 1/x and forget the ln a factor; on the AP exam, that single missed factor costs a point.
- d/dx (ln |x|) = 1/x. The absolute value extends the natural log to negative x. AP Calculus AB and BC both reward the absolute-value form when differentiating ln of an expression that may be negative, particularly inside a related-rates or optimisation problem.
A useful practice is to write each rule in three forms: as a derivative formula, as an antiderivative formula, and as a graph description. For example, d/dx (ln x) = 1/x, ∫ 1/x dx = ln |x| + C, and the graph of ln x is concave down with a vertical asymptote at x = 0. AP exam writers routinely move between these three forms, and the candidates who handle the transitions cleanly are the ones who can move a logarithmic expression from one section of a free-response problem to the next without re-deriving the rule.
Applying the chain rule to e and ln without losing the inner derivative
On the AP exam, the derivative of ex almost never appears by itself. The function is typically composed with something more complicated: e2x, esin x, ex², or even earctan x. The chain rule, d/dx f(g(x)) = f′(g(x)) · g′(x), applies in full, and the trap is that the inner derivative is easy to drop. The rule of thumb I give candidates is: differentiate the outer function first, then multiply by the derivative of whatever is in the exponent. Read that sentence slowly the first time. It is the difference between earning the chain-rule point and losing it.
Consider d/dx (e3x²). The outer function is eu, whose derivative is eu. The inner function is u = 3x², whose derivative is 6x. The answer is 6x · e3x². The 6x sits in front; the exponential form is preserved exactly. A common error is to write the answer as e3x² + 6x, which conflates multiplication of the outer derivative with addition in the exponent. That conflation is one of the most consistent errors graders report in the free-response section.
For logarithmic compositions, the pattern is similar: d/dx (ln f(x)) = f′(x) / f(x). The derivative of the inside goes on top, the inside goes on the bottom, and the natural log disappears entirely. So d/dx (ln(x³ + 1)) = 3x² / (x³ + 1), and d/dx (ln(cos x)) = -sin x / cos x = -tan x. The simplification in the second example is the kind of step that earns an extra method point on a free-response problem, because the AP rubric rewards algebraic work that leads to a closed form.
A practical tip: when the inside of a natural log is a product or quotient, candidates should reach for logarithmic differentiation rather than the quotient rule. Writing ln y = ln(…) turns a quotient of complicated factors into a sum, and the derivative of a sum is easier to manage. This technique is tested directly on the BC exam and indirectly on the AB exam whenever a problem involves differentiating y = (sin x)x or another variable-base, variable-exponent expression. For most candidates reading this, the move to logarithmic differentiation is the single highest-leverage technique in the exponential-logarithmic unit, and it is worth practising until the rewrite feels mechanical.
Derivatives of general bases: ax, xx, and the cases AP exam writers like to test
The general base case, d/dx (ax) = ax · ln a, is the rule that AB and BC candidates most often forget the constant for. The exponential function ax is not its own derivative unless a = e; for any other base, the derivative carries the multiplicative factor ln a. For base 2, ln 2 is approximately 0.693, and the exam routinely uses base 2 in growth-and-decay problems because it ties to the concept of doubling time. For base 10, ln 10 is approximately 2.303, and the exam uses it when a problem states a quantity is growing logarithmically with common log implied.
Variable-base, variable-exponent expressions like xx, (sin x)cos x, or (1 + 1/x)x cannot be differentiated with a single rule. They require logarithmic differentiation: take ln of both sides, differentiate implicitly, then solve for dy/dx. For y = xx, the steps are:
- ln y = x · ln x.
- Differentiate both sides: (1/y) · dy/dx = ln x + x · (1/x) = ln x + 1.
- Solve for dy/dx: dy/dx = y · (ln x + 1) = xx · (ln x + 1).
The same three-step structure works for any expression of the form f(x)g(x). The BC exam tests this technique directly in the free-response section, often as part of a larger problem about limits, where the limit of (1 + 1/n)n as n → ∞ is a classic application. Candidates who can move through the three steps of logarithmic differentiation without freezing are usually the ones who pick up the full method points on those problems.
A second pattern that AP writers like to test is implicit differentiation involving exponentials. The equation y = exy is a typical example. Differentiating implicitly, dy/dx = exy · (y + x · dy/dx), then collecting dy/dx on the left and solving, gives dy/dx = y · exy / (1 - x · exy). The pattern here is that exponentials differentiate to themselves, the chain rule introduces a derivative of the exponent, and the implicit-differentiation step adds the product-rule treatment of x · y. Candidates who keep these three layers separate in their working memory are the ones who finish the problem on the first attempt.
Worked examples in the style of AP free-response questions
Free-response problems on the AP Calculus AB and BC exams are graded against a published rubric, and the rubric rewards specific moves. For exponential and logarithmic derivatives, the moves that earn method points are: writing the derivative rule by name or by formula, applying the chain rule explicitly when the exponent or the argument of the log is a function of x, and simplifying the final answer into a closed form. The three examples below are written in the style of AP free-response items; working through them on paper under timed conditions is the highest-leverage way to internalise the structure.
Example 1 (AB-level). Let f(x) = x · ln(1 + x²). Find f′(x). Solution: f′(x) = ln(1 + x²) + x · (2x / (1 + x²)) = ln(1 + x²) + 2x² / (1 + x²). The product rule produces two terms, and the chain rule inside the second term is the layer AP graders check for. A common error is to write 1/(1 + x²) and forget the 2x from differentiating the inside; that error loses the chain-rule point.
Example 2 (AB-level). Let g(x) = ln(sin x) for 0 < x < π. Find g′(x). Solution: g′(x) = cos x / sin x = cot x. The simplification to cot x is the kind of clean closed form that earns a simplification point on the AP rubric. Candidates who stop at cos x / sin x leave that point on the table.