Bottom × derivative of top, minus top × derivative of bottom, over bottom²: To differentiate a quotient y = u/v: (u/v)' = (u'v − uv') / v².
The order matters — it's u'v minus uv', all over v squared.
IB-style question — a rational function
Differentiate y = (x + 1)/(x − 2).
Step by step
- u = x+1 (u'=1), v = x−2 (v'=1). Apply (u'v − uv')/v².
- Simplify the numerator.
Final answer
dy/dx = −3/(x − 2)².
u'v − uv', not uv' − u'v: Start with the derivative of the top times the bottom — getting the order wrong flips the sign.
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Label, substitute, simplify the numerator: Identify u, v, u', v', substitute into (u'v − uv')/v², then simplify the numerator carefully (signs!).
The denominator stays as v².
IB-style question — quotient
Differentiate y = (2x)/(x² + 1).
Step by step
- u = 2x (u'=2), v = x²+1 (v'=2x).
- Simplify the numerator.
Final answer
dy/dx = (2 − 2x²)/(x² + 1)².
Watch the subtraction: Distribute the minus across the whole uv' term — a sign slip in the numerator is the usual error.
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Same rule, standard derivatives for u' and v': The quotient rule works with any functions — use the standard derivatives for u' and v'. 'Show that' parts ask you to simplify to a given form.
IB-style question — show that
Show that the derivative of y = (ln x)/x is (1 − ln x)/x².
Step by step
- u = ln x (u' = 1/x), v = x (v' = 1).
- Simplify the numerator.
Final answer
dy/dx = (1 − ln x)/x², as required.
'Show that' → end at the given form: For a 'show that', make your simplification arrive exactly at the printed expression.
IB-style question — quotient with a bracket-power
Differentiate y = 10x ⁄ (x² + 3)².
Step by step
- Quotient rule with u = 10x (u′ = 10) and v = (x² + 3)². By the chain rule v′ = 2(x² + 3)(2x) = 4x(x² + 3).
- Cancel one factor of (x² + 3) and simplify the top.
Final answer
y′ = 30(1 − x²) ⁄ (x² + 3)³.