Differentiate powers of \(x\) after rewriting the expression as a sum of terms \(ax^n\) (expand brackets, split fractions, convert roots). Then apply: gradient of tangent/normal, stationary points and their nature, increasing/decreasing intervals, and optimisation.
Derivatives (MM6.1, MM6.2)
- \(\dfrac{d}{dx}x^n = nx^{n-1}\) for any rational \(n\); constants differentiate to \(0\); sums term by term.
- Rewrite first: \(\dfrac{(3x - 2)^2}{x^2} = 9 - 12x^{-1} + 4x^{-2}\); \(\sqrt x = x^{1/2}\); \(\dfrac{1}{\sqrt x} = x^{-1/2}\).
- \(f'(a)\) is the gradient of the tangent at \(x = a\) and the rate of change there; velocity is \(\frac{ds}{dt}\).
- Second derivative \(f''(x)\): positive at a stationary point means a minimum, negative means a maximum, zero is inconclusive (check the sign change of \(f'\)).
Applications (MM6.3, MM8.5)
- Tangent at \((a, f(a))\): \(y - f(a) = f'(a)(x - a)\); normal: gradient \(-\frac{1}{f'(a)}\).
- Stationary points: solve \(f'(x) = 0\). Increasing where \(f'(x) > 0\), decreasing where \(f'(x) < 0\).
- A cubic \(ax^3 + bx^2 + cx + d\) has two turning points iff \(b^2 > 3ac\); it has three real roots iff the local max is above the axis and the local min below.
- Optimisation: express the quantity in one variable, differentiate, solve, confirm max/min, then evaluate.
- \(f'(x) = (x - 2)^2(x + 1)\): the squared factor does not change sign, so \(x = 2\) is a stationary point but not a turning point.
Must-know (no formula booklet)
- \(\frac{d}{dx}x^n = nx^{n-1}\)
- tangent: \(y - y_1 = f'(x_1)(x - x_1)\)
- normal gradient \(= -1/f'(x_1)\)
- \(f'' > 0\) min, \(f'' < 0\) max
- increasing \(\Leftrightarrow f' > 0\)
Classic traps
- Differentiate the whole expression, not a bracket 'in place': \(\frac{d}{dx}(3x - 2)^2 \neq 2(3x - 2)\) without the chain rule (not in the spec - expand instead).
- A local minimum need not be the global minimum: \(x^3 - 3x\) has local min \(-2\) but is unbounded below.
- Minimum distance from a point to a curve: minimise the squared distance to avoid square roots.
Worked example
Q. Find the stationary points of \(y = x^3 - 3x^2 - 9x + 5\).
A. \(y' = 3(x - 3)(x + 1)\); \(y'' = 6x - 6\): maximum \((-1, 10)\), minimum \((3, -22)\).
How it has been examined
October 2018 · Paper 1Question 13graph of f': which point is a local minimum of fquestion papersit online