Picture: four new trig curves, four new slopes: In SL you learned the slopes of sin and cos. HL adds the slopes of the other four trig functions.
Think of each as a curve you could draw on a GDC; the formula below is the gradient at any point. You don't derive these in the exam — you look them up and use them. The job is to recognise the function and quote the right line.
How to remember the signs: The two co-functions (csc and cot) carry a minus sign, exactly like cos x did in SL (d/dx of cos x = −sin x).
So: tan and sec are positive, csc and cot are negative. The 'co' partner always picks up the minus.
IB-style question — read off a derivative
A student is asked for the gradient function of y = tan x.
Write down dy/dx, then find the gradient at x = π/4.
Step by step
- This is the first table entry — quote it directly.
- At x = π/4, sec(π/4) = 1/cos(π/4) = 1/(1/√2) = √2.
- Square it for sec².
Final answer
dy/dx = sec²x, and the gradient at x = π/4 is 2.
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Picture: any base, not just e: In SL the special base was e, where d/dx(eˣ) = eˣ. HL handles any positive base a (like 2ˣ or 10ˣ).
The key idea: every exponential can be written through e, so a factor of ln a appears. The bigger the base, the steeper the curve — and ln a is exactly that steepness factor.
Why ln a shows up: Write aˣ = ex ln a (because a = eln a). Differentiating ex ln a by the chain rule brings down the constant ln a, leaving ex ln a · ln a = aˣ ln a.
Check it works for base e: ln e = 1, so d/dx(eˣ) = eˣ · 1 = eˣ. The SL rule is just the special case a = e.
IB-style question — differentiate a base-2 exponential
Let f(x) = 2ˣ.
Find f′(x), and hence find f′(3). Give your answer to 3 significant figures.
Step by step
- Use d/dx(aˣ) = aˣ ln a with a = 2.
- Substitute x = 3, so 2³ = 8.
- Evaluate (ln 2 ≈ 0.6931).
Final answer
f′(x) = 2ˣ ln 2, and f′(3) = 8 ln 2 ≈ 5.55 (3 s.f.).