IB Maths AI HL Further Differentiation Paper 1 & 2 ~7 min read

Differentiating Special Functions

The power rule handles polynomials, but exams also throw sin, cos, tan, ex and ln x at you. Each has its own standard derivative — most are in the formula booklet — and when the inside is a linear expression like ax + b, you simply multiply by the constant a. Get these results into muscle memory and a whole new class of questions opens up.

📘 What you need to know

Differentiating trig functions

The three basic trig derivatives are given to you in the formula booklet, so the skill is recognising them and handling the inside function. Always work in radians.

Standard trig derivatives y = sin xdydx = cos x  ·  y = cos x → −sin x  ·  y = tan x1cos²x ✓ all three given in the formula booklet
FunctionLinear inside (ax+b)General inside f(x)
sina cos(ax+b)f(x) cos(f(x))
cosa sin(ax+b)f(x) sin(f(x))
tanacos²(ax+b)f(x)cos²(f(x))

🧠 “Sine to cosine, cosine flips sign”

sin differentiates to cos (no sign change); cos differentiates to −sin (a minus appears). The minus lives with the cosine. Whatever’s inside, multiply by its derivative.

⚠ Radians, every time

Differentiating ex & ln x

Exponentials and logarithms have two derivatives worth knowing cold. The exponential is unique — it’s its own derivative.

Exponential & logarithm y = exdydx = ex  ·  y = ln x1x ✗ not in the booklet — learn these two
FunctionLinear inside (ax+b)General inside f(x)
e(…)a e(ax+b)f(x) ef(x)
ln(…)aax+bf(x)f(x)

🤔 Why is the derivative of ln kx still just 1x?

Using the linear rule, ln(kx) differentiates to kkx — and the k‘s cancel, leaving 1x. The constant inside a log simply doesn’t survive differentiation. (It makes sense via log laws too: ln kx = ln k + ln x, and ln k is a constant that differentiates to zero.)

⚠ Two classic slips

Worked examples

WE 1

Differentiate f(x) = sin x

The basic result, straight from the booklet.

f(x) = sin x f′(x) = cos x
WE 2

Differentiate f(x) = cos(5x + 1)

Linear inside: a = 5. Use −a sin(ax+b).

inside = 5x + 1, so a = 5 f′(x) = −5 sin(5x + 1)
WE 3

A curve has equation y = tan(6x2π4). Find the gradient at x = √π2 (exact value).

General inside f(x) = 6x2π4, so f(x) = 12x. Use f(x)cos²(f(x)).

dy/dx = 12xcos²(6x² − π/4) At x = √π/2: 6x² = 6·π4 = 2, so inside = 2π4 = 4 numerator 12x = 12·√π2 = 6√π; cos²(4) = (−1√2)² = 12 dy/dx = 6√π1/2 = 12√π gradient = 12√π
WE 4

A curve is y = e−3x+1 + 2 ln 5x. Find the gradient at x = 2 in the form a + bec.

Differentiate each term: linear exponential (×−3) and the special ln case.

dy/dx = −3e−3x+1 + 2·1x (ln 5x → 1/x, the special b = 0 case) At x = 2: −3e−3(2)+1 + 22 = −3e−5 + 1 gradient = 1 − 3e⁻⁵ (a = 1, b = −3, c = −5)
WE 5

Spot the common slip: differentiate y = ln 7x and y = e4x

Apply the linear-inside rules carefully.

ln 7x → 77x = 1x (the 7’s cancel) e4x → 4e4x (multiply by a = 4) ln 7x → 1/x · e⁴ˣ → 4e⁴ˣ NOT 7/x, and NOT 4x·e^(4x−1).

💡 Top tips

⚠ Common mistakes

Next up — Chain Rule. You’ve already met the pattern: when the inside of sin, cos, e or ln is a function of x, you multiply by its derivative f(x). That “multiply by the derivative of the inside” move is the chain rule — the next topic makes it explicit and shows how to handle any composite “function of a function”, not just these special ones.

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