Picture: nothing new about the rules — only the parts: You already know the product rule (uv)′ = u′v + uv′ and the quotient rule. HL just lets the pieces be tan, sec, arctan, aˣ and friends.
The whole skill is: spot u and v, differentiate each using the 5.15 table, then slot them in. No new rule to learn — just new derivatives to feed in.
IB-style question — product rule with arctan
Let f(x) = x² arctan x.
Find f′(x).
Step by step
- Choose the two factors: u = x² and v = arctan x.
- Differentiate each (the second uses the new table).
- Apply (uv)′ = u′v + uv′.
- Tidy the second term.
Final answer
f′(x) = 2x arctan x + x²/(1 + x²).
IB-style question — product rule with sec
Let f(x) = eˣ sec x.
Find f′(x) and factorise your answer.
Step by step
- u = eˣ (derivative eˣ), v = sec x (derivative sec x tan x).
- Apply the product rule.
- Both terms share eˣ sec x — factor it out.
Final answer
f′(x) = eˣ sec x (1 + tan x).
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Picture: differentiate the outside, then multiply by the inside's slope: The chain rule says: differentiate the outer function (using the 5.15 table), keep the inner function inside, then multiply by the inner's derivative.
For example arctan(3x): outer derivative 1/(1 + (inner)²) with inner = 3x, times the inner's slope 3.
IB-style question — chain rule with arctan
Differentiate y = arctan(3x).
Step by step
- Outer is arctan(·), whose derivative is 1/(1 + (·)²). Keep the inner 3x inside.
- Multiply by the inner's derivative, d/dx(3x) = 3.
- Simplify (note (3x)² = 9x²).
Final answer
dy/dx = 3/(1 + 9x²).
IB-style question — chain rule with tan
Differentiate y = tan(x²).
Step by step
- Outer is tan(·), derivative sec²(·). Keep the inner x² inside.
- Multiply by d/dx(x²) = 2x.
- Write the constant factor in front.
Final answer
dy/dx = 2x sec²(x²).