A curve has no single gradient, but at each point it has one: the gradient of the tangent there. Squeeze a chord onto the point and its gradient settles on that value, which is what the derivative measures.
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
Section 01
From chords to a tangent.
In the lab, fix a point on the curve and drag the second point of the chord towards it. Watch the chord's gradient in the readout: it changes less and less, and closes in on one number. That limiting value is the gradient of the curve at the point.
Chord gradient: for points with \(x\)-coordinates \(a\) and \(a + h\), the gradient is \(\dfrac{f(a+h) - f(a)}{h}\).
Shrink \(h\): as \(h \to 0\) the chord turns into the tangent at \(x = a\).
The limit of the chord gradients is the gradient of the curve at that point. Cambridge expects only this informal idea of a limit, not a formal proof from first principles.
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 \(6\).
Worked example Chords on a cubic
Find the gradient of the chord joining the points with \(x\)-coordinates \(2\) and \(2 + h\) on \(y = x^{3}\). Hence find the gradient of the curve at \((2, 8)\).
As \(h \to 0\), the terms \(6h\) and \(h^{2}\) vanish.
Answer: Chord gradient \(12 + 6h + h^{2}\); gradient of the curve at \((2, 8)\) is \(12\) (Fig. 1).
Examiner trap
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 \(h \to 0\).
Section 02
The derivative and its notation.
Doing the chord calculation at a general point \(x\) gives a formula for the gradient at every point: the derivative. For powers of \(x\) the pattern is short enough to use directly.
Notation
Meaning
\(\dfrac{dy}{dx}\) or \(f'(x)\)
first derivative: the gradient of \(y = f(x)\) at \(x\)
\(\dfrac{d^{2}y}{dx^{2}}\) or \(f''(x)\)
second derivative: differentiate \(\dfrac{dy}{dx}\) again
\(y = x^{n}\)
\(\dfrac{dy}{dx} = nx^{n-1}\)
\(y = k\,f(x)\), \(y = f(x) \pm g(x)\)
\(k\,f'(x)\) and \(f'(x) \pm g'(x)\): term by term
\(y = c\) (a constant)
\(\dfrac{dy}{dx} = 0\): a horizontal line has zero gradient
Worked example Term by term
Given \(y = 4x^{3} - 5x^{2} + 7x - 2\), find \(\dfrac{dy}{dx}\) and \(\dfrac{d^{2}y}{dx^{2}}\), and the gradient of the curve at \(x = 1\).
\(\dfrac{d^{2}y}{dx^{2}}\) is NOT \(\left(\dfrac{dy}{dx}\right)^{2}\). For \(y = x^{3}\), \(\dfrac{d^{2}y}{dx^{2}} = 6x\), but \(\left(\dfrac{dy}{dx}\right)^{2} = 9x^{4}\).
Stage 1 · Learn
Check your understanding
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On \(y = x^{2}\), what is the gradient of the chord joining the points with \(x\)-coordinates \(1\) and \(1 + h\)?
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Find \(\dfrac{dy}{dx}\) when \(y = x^{5}\).
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\(f(x) = x^{3}\). Which is \(f''(x)\)?
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Find the gradient of \(y = 3x^{2}\) at \(x = -2\).
Section 03
Using the gradient.
Once you have \(\dfrac{dy}{dx}\), you can read off the gradient at any point, or run it backwards: set \(\dfrac{dy}{dx}\) equal to a value and solve for \(x\).
Worked example Where is the gradient a given value?
Find the coordinates of the points on \(y = x^{3} - 12x\) where the gradient is zero.
\(\dfrac{dy}{dx} = 3x^{2} - 12\).
Set \(3x^{2} - 12 = 0\), so \(x^{2} = 4\) and \(x = 2\) or \(x = -2\).
Find \(y\) from the curve, not from \(\dfrac{dy}{dx}\): \(x = 2\) gives \(8 - 24 = -16\); \(x = -2\) gives \(-8 + 24 = 16\).
Answer: \((2, -16)\) and \((-2, 16)\)
Examiner trap
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.
Stage 2 · Exam
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The points \(A\) and \(B\) on \(y = x^{3}\) have \(x\)-coordinates \(2\) and \(2 + h\). Which expression is the gradient of the chord \(AB\)?
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\(f(x) = 2x^{4} - 3x^{2} + 5x\). Find \(f'(-1)\).
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The curve \(y = x^{3} - 6x^{2} + 5\) has gradient \(-9\) at two points. Find their \(x\)-coordinates.
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Given \(y = x^{4} - 2x^{3}\), find the value of \(\dfrac{d^{2}y}{dx^{2}}\) when \(x = 2\).
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Gradients of curves, found.
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