IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~11 min read

Differentiating Exponential & Logarithmic Functions

The natural cases d/dx[e^x] = e^x and d/dx[ln x] = 1/x are the simplest. For general bases, the derivatives pick up an extra ln a factor: d/dx[a^x] = a^x ln a, and d/dx[log_a x] = 1/(x ln a). Both proved via logarithmic differentiation — the technique of taking ln of both sides and differentiating implicitly.

📘 What you need to know

Special cases (formula booklet): d/dx[e^x] = e^x; d/dx[ln x] = 1/x.
General cases (formula booklet): d/dx[a^x] = a^x · ln a; d/dx[log_a x] = 1/(x · ln a).
The “extra” ln a: when a = e, ln a = 1 and the formulas collapse to the natural case. The general formula always carries this factor.
Geometric meaning of ln a: it’s the slope of y = a^x at x = 0. Different bases give different “steepness” at the curve’s intersection with the y-axis.
Chain rule for linear arguments: d/dx[a^px+q] = p · a^px+q · ln a. For logs: d/dx[log_a(px+q)] = p / ((px+q) · ln a).
Logarithmic differentiation: take natural log of both sides, simplify with log laws, differentiate implicitly. Required for things like y = x^x (both base and exponent are variable).
Log laws can simplify before differentiating — e.g., ln(x²/sin x) = 2 ln x − ln(sin x) makes the derivative cleaner.
Domain reminder: a^x requires a > 0 (and usually a ≠ 1); log_a x requires x > 0.

The four derivatives — special and general

Formula booklet — exp and log derivatives ddx[e^x] = e^x | ddx[ln x] = 1x | ddx[a^x] = a^x ln a | ddx[log_a x] = 1x ln a

The general formulas reduce to the natural ones when a = e, because ln(e) = 1. Geometrically, the “extra” ln a factor controls how steeply y = a^x rises:

Both curves pass through (0, 1) — because a⁰ = 1 for every base. At that point, the BLUE tangent (slope ≈ 0.69) is shallower than the ORANGE tangent (slope ≈ 1.61). The slopes are exactly ln 2 and ln 5 — matching the formula d/dx[a^x]|_x=0 = a⁰ · ln a = ln a.

Why ln a appears: starting from y = a^x, take ln of both sides → ln y = x ln a. Differentiate implicitly: (1/y) · dy/dx = ln a. So dy/dx = y · ln a = a^x · ln a. The ln a comes from the constant in front of x.

Chain rule for compound arguments

Exponential — argument u(x)

ddx[a^u] = a^u · ln a · dudx

multiply by ln a AND by the inner derivative

Logarithm — argument u(x)

ddx[log_a u] = du/dxu · ln a

inner derivative in numerator; u and ln a in denominator

🧭 Recipe — differentiate any exp or log expression

Identify the function type: e^x, a^x, ln x, or log_a x. Note whether the argument is just x or something more complex.
Quote the formula-booklet derivative for the base form. Remember: general bases pick up a ln a factor.
Apply chain rule if the argument u is not just x: for exp → multiply by du/dx; for log → put du/dx in the numerator.
For products, quotients, or composite expressions, apply the relevant rule (product / quotient / chain) with the formula-booklet derivative plugged in.
For expressions like x^x or (sin x)^x where both base and exponent depend on x, use LOGARITHMIC DIFFERENTIATION: take ln of both sides, simplify, then differentiate implicitly.

Worked examples

WE 1

Prove d/dx[a^x] = a^x ln a via logarithmic differentiation

Show, using logarithmic differentiation, that the derivative of y = a^x is a^x · ln a, where a is a positive constant.

Step 1 — take natural log of both sides y = a^x ln y = ln(a^x) = x · ln a (power-law for logs) Step 2 — differentiate both sides with respect to x (implicit) d/dx[ln y] = d/dx[x · ln a] (1/y) · (dy/dx) = ln a (ln a is a CONSTANT, not differentiated) Step 3 — solve for dy/dx dy/dx = y · ln a Step 4 — substitute back y = a^x dy/dx = a^x · ln a ✓ ∴ d/dx[a^x] = a^x · ln a ln a is a fixed number (since a is constant), so when differentiating x · ln a it acts just like a coefficient. The whole proof rests on log laws + implicit differentiation.

WE 2

Chain rule on an exponential with NONLINEAR argument

Find dydx for y = 2^{x² + 1}.

Step 1 — identify the inner function u u = x² + 1 ⟹ du/dx = 2x Step 2 — apply chain rule with d/dx[a^u] = a^u · ln a · du/dx dy/dx = 2^(x² + 1) · ln 2 · d/dx[x² + 1] = 2^(x² + 1) · ln 2 · 2x Step 3 — tidy up = 2x · 2^(x² + 1) · ln 2 dy/dx = 2x · 2^(x² + 1) · ln 2 three multiplicative factors: the original exponential 2^(x² + 1), the ln 2 constant, and the inner derivative 2x. Don’t drop any of them.

WE 3

Chain rule on a logarithm with linear argument

Find dydx for y = log₃(4x + 1).

Step 1 — identify u and du/dx u = 4x + 1 ⟹ du/dx = 4 Step 2 — apply chain rule with d/dx[log_a u] = (du/dx) / (u · ln a) dy/dx = 4 / ((4x + 1) · ln 3) dy/dx = 4(4x + 1) · ln 3 the inner derivative (4) lives in the NUMERATOR; the inner expression (4x + 1) and ln 3 both live in the DENOMINATOR. A common mistake is to put ln 3 in the wrong place.

WE 4

Product rule with general-base exponential

Find dydx for y = x · 2^x.

Step 1 — identify the two factors u = x, u’ = 1 v = 2^x, v’ = 2^x · ln 2 Step 2 — apply product rule: (uv)’ = u’v + uv’ dy/dx = (1)(2^x) + (x)(2^x · ln 2) = 2^x + x · 2^x · ln 2 Step 3 — factor 2^x for cleaner form = 2^x · (1 + x · ln 2) dy/dx = 2^x (1 + x ln 2) factor out the common 2^x at the end if asked for “simplified” form. The natural exp case y = x · e^x would give e^x(1 + x) — much tidier because ln(e) = 1.

WE 5

Logarithmic differentiation on y = x^x

Find dydx for y = x^x, where x > 0.

Why neither power rule nor exp rule alone work y = x^x has VARIABLE base AND VARIABLE exponent power rule needs constant exponent; exp rule needs constant base Step 1 — take natural log of both sides ln y = ln(x^x) = x · ln x (power-law for logs) Step 2 — differentiate both sides implicitly with respect to x d/dx[ln y] = (1/y) · (dy/dx) (chain rule on left) d/dx[x · ln x] = 1 · ln x + x · (1/x) = ln x + 1 (product rule on right) ⟹ (1/y) · (dy/dx) = ln x + 1 Step 3 — solve for dy/dx dy/dx = y · (1 + ln x) Step 4 — substitute back y = x^x dy/dx = x^x · (1 + ln x) dy/dx = x^x (1 + ln x) the technique works for any “tower” expression like x^x, x^(2x), (sin x)^x — take ln of both sides, use log laws to bring the exponent down to a coefficient, then differentiate implicitly.

WE 6

Equation of tangent to y = x ln x at x = 1

Find the equation of the tangent to the curve y = x ln x at the point where x = 1.

Step 1 — find the y-coordinate y at x = 1: (1) · ln(1) = 1 · 0 = 0 point of tangency: (1, 0) Step 2 — find dy/dx using product rule u = x, u’ = 1; v = ln x, v’ = 1/x dy/dx = 1 · ln x + x · (1/x) = ln x + 1 Step 3 — gradient at x = 1 dy/dx |_{x = 1} = ln 1 + 1 = 0 + 1 = 1 Step 4 — tangent y – y₀ = m(x – x₀) y – 0 = 1 · (x – 1) y = x – 1 tangent: y = x − 1 at x = 1 (where the curve crosses the x-axis), the tangent is y = x − 1 — a straight line of slope 1. Geometrically, the curve y = x ln x crosses the x-axis at x = 1 with gradient exactly 1.

💡 Top tips

Quote the formula booklet for ALL four core derivatives. Even d/dx[ln x] = 1/x — examiners notice the citation.
The ln a factor never disappears for general bases. Forgetting it is the single most-common error in this topic.
For logs with compound arguments, use log laws to simplify FIRST when possible: log_a(x²y) = 2 log_a x + log_a y is much easier to differentiate.
Logarithmic differentiation is the right tool whenever you have a variable raised to a variable power (like x^x), or a really messy product/quotient where ln converts it to a sum.
e^x is self-derivative — d/dx[e^x] = e^x. Unique among all functions. That’s WHY e is called the “natural” base.

⚠ Common mistakes

Forgetting the ln a factor on general-base derivatives. d/dx[2^x] = 2^x · ln 2, NOT just 2^x. (The natural-base shortcut only works when a = e.)
Confusing the power rule with the exp rule: d/dx[x³] = 3x² (power rule, fixed exponent), but d/dx[3^x] = 3^x · ln 3 (exp rule, fixed base).
Putting ln a in the wrong place for logs: d/dx[log₅ x] = 1/(x · ln 5), NOT (ln 5)/x or 1/(x · log 5).
Treating ln a as a function of x. ln a is a CONSTANT (since a is a constant), so it differentiates to 0 — not to 1/a.
Trying to apply the power rule to x^x: that gives x · x^x−1 = x^x, which is WRONG. The actual derivative is x^x(1 + ln x), found via logarithmic differentiation.

That’s the full Further Differentiation chapter — first principles, inverse functions, implicit differentiation, related rates, reciprocal trig, inverse trig, and now exp/log. Up next: Further Integration. With every derivative pattern in your toolkit, you’ll meet their integration counterparts (integration by substitution and integration by parts), and HL-only techniques like partial fractions. Every derivative becomes an integral you can reverse — and that’s the bridge to the Fundamental Theorem of Calculus.

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