Everything from the 4 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
The gradient of a curve at a point is the gradient of the tangent there, the limit of the gradients of chords as the second point slides onto the first. For \(y = x^{n}\), \(\dfrac{dy}{dx} = nx^{n-1}\): bring the power down, then reduce the power by one.
Fig. 1: on y = x³ the chord PQ from x = 2 to x = 2 + h has gradient 12 + 6h + h² (15.25 when h = 0.5, as drawn); as h shrinks to 0 it tends to 12, the gradient of the tangent at P
Worked example: Chords on a quadratic
The points \(P(3, 9)\) and \(Q(3 + h, (3+h)^{2})\) lie on \(y = x^{2}\). Find the gradient of the chord \(PQ\) in terms of \(h\), and deduce the gradient of the curve at \(P\).
Answer: Chord gradient \(6 + h\); the gradient of the curve at \(P\) is \(6\).
Differentiate, then substitute or solve.
Chord: gradient \(= \dfrac{f(a+h) - f(a)}{h}\); simplify, cancel \(h\), then let \(h \to 0\).
Power rule: \(x^{n} \to nx^{n-1}\); constants vanish; work term by term.
Gradient at a point: substitute \(x\) into \(\dfrac{dy}{dx}\).
Point with a given gradient: solve \(\dfrac{dy}{dx} = m\), then use the curve for \(y\).
Second derivative: differentiate \(\dfrac{dy}{dx}\) again.
Watch out!
Do not put \(h = 0\) into \(\dfrac{12h + 6h^{2} + h^{3}}{h}\) straight away: that gives \(\dfrac{0}{0}\), which means nothing. Cancel the \(h\) first, THEN let \(h \to 0\).
\(\dfrac{d^{2}y}{dx^{2}}\) is NOT \(\left(\dfrac{dy}{dx}\right)^{2}\). For \(y = x^{3}\), \(\dfrac{d^{2}y}{dx^{2}} = 6x\), but \(\left(\dfrac{dy}{dx}\right)^{2} = 9x^{4}\).
After solving \(\dfrac{dy}{dx} = 0\) for \(x\), substitute into the equation of the CURVE to get \(y\). Putting \(x\) back into \(\dfrac{dy}{dx}\) just gives 0 again.
Write every term as \(kx^{n}\) before differentiating. For a function of a function, \(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\): differentiate the outside leaving the inside alone, then multiply by the derivative of the inside. So \(\dfrac{d}{dx}(ax + b)^{n} = an(ax + b)^{n-1}\).
Fig. 1: y = (3x − 1)⁵ is a chain of two simple functions; multiply the two rates of change to get dy/dx
Worked example: Roots and reciprocals
Given \(y = 6\sqrt{x} + \dfrac{8}{x^{2}}\), find \(\dfrac{dy}{dx}\) and the gradient of the curve at \(x = 4\).
Rewrite as powers: \(y = 6x^{\frac{1}{2}} + 8x^{-2}\).
Linear inside: \(\dfrac{d}{dx}(ax + b)^{n} = an(ax + b)^{n-1}\).
Tidy: write negative and fractional powers back as fractions and roots if asked.
Watch out!
\(\dfrac{1}{2x^{2}}\) is \(\tfrac{1}{2}x^{-2}\), NOT \(2x^{-2}\): only the \(x^{2}\) moves up with a negative power; the 2 stays in the denominator as \(\tfrac{1}{2}\).
The most common lost mark: forgetting to multiply by the derivative of the inside. \(\dfrac{d}{dx}(3x - 1)^{5}\) is \(15(3x - 1)^{4}\), not \(5(3x - 1)^{4}\). Watch the sign too when the inside is something like \(4 - x^{2}\).
Do not expand \((2x - 3)^{4}\) before differentiating unless you have to. It is slow, and a single slip in the binomial expansion costs every later mark.
At a point where the gradient is \(m\), the tangent has gradient \(m\) and the normal has gradient \(-\dfrac{1}{m}\). \(f\) is increasing where \(f'(x) > 0\) and decreasing where \(f'(x) < 0\). Connected rates: \(\dfrac{dA}{dt} = \dfrac{dA}{dr} \times \dfrac{dr}{dt}\).
Fig. 1: the tangent (gradient 4) and the normal (gradient −1/4) to y = x³ − 2x² + 1 at (2, 1), drawn to equal scales so the right angle is true
Worked example: Tangent and normal to a cubic
Find the equations of the tangent and the normal to \(y = x^{3} - 2x^{2} + 1\) at the point where \(x = 2\).
Gradient: substitute \(x\) into \(\dfrac{dy}{dx}\); normal uses \(-\dfrac{1}{m}\).
Line: \(y - y_{1} = m(x - x_{1})\); give it in the form asked.
Increasing/decreasing: factorise \(f'(x)\), sketch it, read off where it is positive or negative.
Connected rates: know one rate, want another, link with a formula, differentiate, then multiply or divide.
Watch out!
The normal gradient is \(-\dfrac{1}{m}\): flip AND change sign. With \(m = 4\), \(-4\) and \(\tfrac{1}{4}\) are both wrong. And never use \(\dfrac{dy}{dx}\) as a formula in the line equation: substitute to get a number first.
For a positive quadratic \(f'(x)\), the negative part is between the roots, not outside them. Sketch the parabola of \(f'(x)\) before writing the inequality.
Differentiate the formula BEFORE you substitute the particular value. Putting \(r = 5\) into \(V = \tfrac{4}{3}\pi r^{3}\) first turns it into a constant, whose derivative is 0.
Stationary points are where \(\dfrac{dy}{dx} = 0\). If \(\dfrac{d^{2}y}{dx^{2}} < 0\) there, the point is a maximum; if \(\dfrac{d^{2}y}{dx^{2}} > 0\), a minimum. If \(\dfrac{d^{2}y}{dx^{2}} = 0\), check the sign of \(\dfrac{dy}{dx}\) on either side instead.
Fig. 1: y = x³ − 3x² − 9x + 5 has a maximum at (−1, 10) and a minimum at (3, −22); the tangent is horizontal at both
Worked example: Locate and classify
Find the stationary points of \(y = x^{3} - 3x^{2} - 9x + 5\) and determine their nature.
\(\dfrac{d^{2}y}{dx^{2}} = 6x - 6\): at \(x = -1\) it is \(-12 < 0\), a maximum; at \(x = 3\) it is \(12 > 0\), a minimum.
Answer: Maximum \((-1, 10)\); minimum \((3, -22)\) (Fig. 1).
Gradient zero, then test the nature.
Find: solve \(\dfrac{dy}{dx} = 0\); get \(y\) from the curve.
Second derivative: negative means maximum, positive means minimum.
If it is zero: check the sign of \(\dfrac{dy}{dx}\) just either side.
Sketch: stationary points, axis crossings, and the behaviour for large \(|x|\).
Optimise: one-variable formula, differentiate, solve, justify max or min, answer the question asked.
Watch out!
A question that says 'find the stationary points' wants both coordinates. Leaving the answer as \(x = -1\) and \(x = 3\) loses the accuracy marks.
\(\dfrac{d^{2}y}{dx^{2}} = 0\) does NOT tell you the nature. Go back to the sign of \(\dfrac{dy}{dx}\) either side of the point, choosing \(x\)-values close to it.
Answer the question asked: 'find the maximum volume' wants \(V = 128\), not \(x = 2\). And always show that the point IS a maximum, with the second derivative or a sign check.