IB Maths AA HL
Topic 5 — Calculus
Paper 1 & 2
~7 min read
Differentiating Trig, Exp & Log Functions
Three new families of derivatives to memorise: sin/cos/tan, ex, and ln x. The pattern for each is short — and once you’ve got the linear-inside (ax + b) shortcut, you can differentiate things like sin(3x + 2) or e5x − 1 in one line. For trig, your GDC must be in radians.
📘 What you need to know
- Trig basics: d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(tan x) = sec²x.
- Exp basic: d/dx(ex) = ex. The only function whose derivative is itself.
- Log basic: d/dx(ln x) = 1/x for x > 0.
- Linear inside (ax + b): multiply the derivative by a. E.g. d/dx(sin(ax + b)) = a cos(ax + b).
- ln(ax) derivative = 1/x (the a cancels — it does NOT give a/x).
- ekx derivative = k ekx (NOT kx ekx − 1 — that’s the power-rule mistake).
- Radians ONLY for trig calculus. Set your GDC to radians before any trig differentiation question.
- For more general inside f(x): chain rule (next note). The linear-inside rule is the chain-rule special case for ax + b.
Trig derivatives
sin x
d/dx (sin x) = cos x
sine flips to cosine
cos x
d/dx (cos x) = −sin x
cosine flips to −sine (note minus)
tan x
d/dx (tan x) = sec²x
where sec x = 1/cos x
linear inside (ax + b)
d/dx sin(ax + b) = a cos(ax + b)
multiply by the coefficient of x
Radians warning: all calculus rules above only work in radians. If your GDC is in degrees you’ll get garbage. Switch mode the moment you see sin/cos/tan in a derivative question.
ex and ln x derivatives
Core results
d/dx (ex) = ex | d/dx (ln x) = 1x (x > 0)
Linear inside
d/dx (eax + b) = a eax + b d/dx (ln(ax + b)) = aax + b
The two traps to lock in: ln(kx) differentiates to 1/x not k/x (the a‘s cancel — substitute b = 0 in the formula above, you get a/(ax) = 1/x). And ekx differentiates to k·ekx, never kx·ekx − 1 (that’s the power rule — wrong family of functions).
Summary table
| Function y | Derivative dy/dx | Linear-inside version (ax + b) |
|---|
| sin x | cos x | a cos(ax + b) |
| cos x | −sin x | −a sin(ax + b) |
| tan x | sec²x | a sec²(ax + b) |
| ex | ex | a eax + b |
| ln x | 1/x | a/(ax + b) |
🧭 Recipe — differentiate trig/exp/log
- Identify which standard form the function fits (sin, cos, tan, e□, ln □).
- Check if the inside is linear (ax + b) — if so, apply the standard derivative and multiply by a.
- Set GDC to radians if any sin/cos/tan appears.
- Evaluate at the given x-value if a numeric gradient is asked for.
- Express in the requested form (exact, simplified, or a + b·ec).
Worked examples
WE 1Find f′(x) for two trig functions
Find f′(x) for: (a) f(x) = cos x; (b) f(x) = tan(7x − 2).
(a) Standard derivative of cos
f′(x) = −sin x
(b) tan with linear inside (a = 7)
d/dx [tan(ax + b)] = a sec²(ax + b)
f′(x) = 7 sec²(7x − 2)
(a) f′(x) = −sin x | (b) f′(x) = 7 sec²(7x − 2)
don’t forget the minus on cos derivative — it’s the most-missed sign in calculus
WE 2Gradient of a trig curve at a point — exact answer
A curve has equation y = cos(2x² − π/4). Find the gradient of the tangent at the point where x = √π / 2. Give your answer as an exact value.
Identify form: cos(f(x)) where f(x) = 2x² − π/4
d/dx [cos(f(x))] = −f′(x) sin(f(x))
f′(x) = 4x
dy/dx = −4x sin(2x² − π/4)
Substitute x = √π/2
2x² = 2 · π/4 = π/2
2x² − π/4 = π/2 − π/4 = π/4
sin(π/4) = √2/2
dy/dx = −4 · (√π/2) · (√2/2) = −√2 · √π = −√(2π)
Gradient = −√(2π)
at HL the inside is rarely just ax + b — chain rule (next note) gives the f′(x)·standard pattern
WE 3Three exp/log derivatives — basic patterns
Find dy/dx for: (a) y = e4x − 3; (b) y = ln(7x); (c) y = ln(2x + 5).
(a) e^(ax + b) → a e^(ax + b)
dy/dx = 4 e^(4x − 3)
(b) ln(ax) — special case b = 0, a’s cancel
dy/dx = 1/x (NOT 7/x)
(c) ln(ax + b) → a/(ax + b)
dy/dx = 2/(2x + 5)
(a) 4 e^(4x − 3) | (b) 1/x | (c) 2/(2x + 5)
part (b) is the big trap — write ln(7x) = ln 7 + ln x to see why the 7 disappears
WE 4Combined exp + log gradient — exact form
A curve has equation y = e2x + 3 − 4 ln(2x). Find the gradient at x = 1, giving your answer in the form a + b ec where a, b, c are integers.
Differentiate term by term
d/dx [e^(2x + 3)] = 2 e^(2x + 3)
d/dx [4 ln(2x)] = 4 · (1/x) = 4/x (a’s cancel)
dy/dx = 2 e^(2x + 3) − 4/x
Substitute x = 1
dy/dx = 2 e^5 − 4 = −4 + 2 e^5
Gradient at x = 1 is −4 + 2 e⁵ → a = −4, b = 2, c = 5
your GDC will give ≈ 292.83 — useful as a sanity check for the exact form
WE 5Equation of tangent — trig curve at exact x-value
Find the equation of the tangent to y = sin(2x) at x = π/6. Give your answer as an exact equation.
Find y at x = π/6
y = sin(2 · π/6) = sin(π/3) = √3/2
point is (π/6, √3/2)
Find gradient: d/dx [sin(2x)] = 2 cos(2x)
m = 2 cos(π/3) = 2 · (1/2) = 1
Apply point-slope form
y − √3/2 = 1 · (x − π/6)
y = x − π/6 + √3/2
Tangent: y = x − π/6 + √3/2
trig calculus = radians ALWAYS — π/6 means radians, never 30°
WE 6Real-world: bacterial growth rate
A bacterial population is modelled by N(t) = 500 e0.04t where t is hours after the start of the experiment, t ≥ 0.
(a) Find dN/dt. (b) Find the rate of growth at t = 10 hours, to 2 d.p. (c) Find, to 2 d.p., the time at which the rate of growth is 30 bacteria per hour.
(a) Differentiate: e^(at) → a e^(at)
dN/dt = 500 · 0.04 · e^(0.04t) = 20 e^(0.04t)
(b) Substitute t = 10
dN/dt = 20 e^(0.4) = 20 · 1.4918… ≈ 29.84
≈ 29.84 bacteria per hour
(c) Solve dN/dt = 30
20 e^(0.04t) = 30
e^(0.04t) = 1.5
0.04t = ln(1.5)
t = ln(1.5)/0.04 = 25 ln(1.5) ≈ 10.14
(a) dN/dt = 20 e^(0.04t); (b) ≈ 29.84 bact/h; (c) t ≈ 10.14 hours
dN/dt is the rate of change of population — it has units bacteria per hour
💡 Top tips
- GDC into radians the second you see sin/cos/tan in a calculus question — most exam mistakes come from leaving it in degrees.
- ln(kx) differentiates to 1/x, never k/x. Quick check: ln(kx) = ln k + ln x, and ln k is a constant with derivative 0.
- ekx differentiates to k·ekx. Power rule does NOT apply — exponentials live in a different rule family.
- For exact answers, use exact π/6, π/4, π/3 values: sin(π/6)=½, cos(π/6)=√3/2, sin(π/4)=cos(π/4)=√2/2, sin(π/3)=√3/2, cos(π/3)=½, tan(π/4)=1.
- GDC d/dx gives a numeric gradient — useful to verify your exact answer (e.g. WE 4: 2e⁵ − 4 ≈ 292.83).
⚠ Common mistakes
- Forgetting the minus on d/dx(cos x) = −sin x — the most missed sign in trig calculus.
- Power rule on ekx: writing kx·ekx − 1 is wrong. Exponentials don’t lose their power on differentiation.
- ln(kx) → k/x: nope. The k cancels. Answer is just 1/x.
- Degrees instead of radians: sin(30) in degrees gives 0.5, but in calculus you need sin(π/6) in radians = 0.5. Same number, very different mode.
- Forgetting to multiply by a for linear-inside cases: d/dx sin(3x) is 3 cos(3x), not just cos(3x).
Up next: Chain Rule. The general tool for differentiating composite functions — “function of a function” — like sin(x² + 1), ecos x, or ln(3x² − 5x). The linear-inside shortcut you’ve used here is the chain rule’s special case (a constant inside derivative). After chain comes product and quotient rules — the full toolkit for differentiating anything HL throws at you.
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