Extended. Drawing tangents gives an estimate of a curve's gradient. Differentiation gives the exact gradient at every point at once, from one short rule.
If \(y = ax^{n}\) then \(\dfrac{dy}{dx} = anx^{n-1}\): multiply by the power, then take one off the power. A constant term differentiates to 0. Substitute an \(x\)-value into \(\dfrac{dy}{dx}\) to get the gradient there.
Fig. 1: on y = x², the chord from P(2, 4) to Q(3, 9) has gradient 5; as Q slides towards P the chord turns into the tangent at P, gradient 4
Section 01
From chord to tangent.
In the lab, drag P and Q onto the curve and read the gradient of the chord PQ. Then move Q closer and closer to P. The chord's gradient settles on one value: the gradient of the tangent at P.
Q on \(y = x^2\)
Chord PQ from P(2, 4)
Gradient
\((3, 9)\)
\(\dfrac{9 - 4}{3 - 2}\)
5
\((2.5, 6.25)\)
\(\dfrac{6.25 - 4}{0.5}\)
4.5
\((2.1, 4.41)\)
\(\dfrac{4.41 - 4}{0.1}\)
4.1
\((2.01, 4.0401)\)
\(\dfrac{0.0401}{0.01}\)
4.01
Worked example Spotting the gradient function
Using chords and tangents on \(y = x^2\), the gradient at \(x = 2\) is 4. The same method gives gradient 2 at \(x = 1\) and 6 at \(x = 3\). Find a rule.
List the results: \(x = 1 \to 2\), \(x = 2 \to 4\), \(x = 3 \to 6\).
Each gradient is double the \(x\)-value.
So the gradient of \(y = x^2\) at any point is \(2x\).
Answer: Gradient of \(y = x^2\) is \(2x\): this is its gradient function, \(\dfrac{dy}{dx} = 2x\).
Examiner trap
A tangent touches the curve at one point and has the same steepness there. A chord joins two points of the curve and only approximates the gradient: do not stop at the chord.
Section 02
Multiply by the power, reduce it by one.
Open the gradient function tab. Drag the point along the curve and watch the gradient graph being drawn underneath: for \(y = x^2\) it is a straight line, for \(y = x^3\) a parabola. The rule below predicts both.
\(y\)
\(\dfrac{dy}{dx}\)
Why
\(x^3\)
\(3x^2\)
power 3 comes down, power drops to 2
\(5x^4\)
\(20x^3\)
\(5 \times 4 = 20\), power drops to 3
\(7x\)
\(7\)
\(7x^1 \to 7 \times 1 \times x^0 = 7\)
\(9\)
\(0\)
a constant does not change, so its gradient is 0
Worked example A sum of terms
Find \(\dfrac{dy}{dx}\) when \(y = 5x^3 - 4x^2 + 7\).
(a) \(\tfrac{1}{2} \times 4 = 2\), and the power drops to 3: \(\dfrac{dy}{dx} = 2x^3\).
(b) Brackets must be expanded first: \((x - 4)(2x + 1) = 2x^2 + x - 8x - 4 = 2x^2 - 7x - 4\).
Now differentiate: \(4x - 7\).
Answer: (a) \(2x^3\) (b) \(4x - 7\)
Examiner trap
Do not differentiate each bracket and multiply: for \((x - 4)(2x + 1)\) that gives \(1 \times 2 = 2\), which is wrong. Expand first, then differentiate term by term.
Stage 1 · Learn
Check your understanding
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Quick check+10 XP
Find \(\dfrac{dy}{dx}\) when \(y = x^5\).
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Find \(\dfrac{dy}{dx}\) when \(y = 7x\).
Quick check+10 XP
What does a constant term, such as \(+9\), become when you differentiate?
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What is the gradient of \(y = x^2\) at \(x = 3\)?
Section 03
Gradient at a point.
\(\dfrac{dy}{dx}\) is a formula. Substitute an \(x\)-value to find the gradient at that point, or set it equal to a number to find where the curve has that gradient.
Worked example Gradient at a given point
Find the gradient of \(y = x^3 - 2x + 1\) at the point \((2, 5)\).
Substitute into \(\dfrac{dy}{dx}\) for the gradient, but into \(y\) for the coordinate. Putting \(x = 2\) into \(y = x^3 - 2x + 1\) gives 5, the height of the point, not its gradient.
Stage 2 · Exam
Exam-style questions
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\(y = 4x^3 - 3x^2 + 5\). Find \(\dfrac{dy}{dx}\).
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Find the gradient of the curve \(y = x^3 - 2x + 1\) at the point \((2, 5)\).
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\(y = (x + 3)(2x - 1)\). Find \(\dfrac{dy}{dx}\).
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Find the coordinates of the point on \(y = x^2 - 4x + 3\) where the gradient is 6.
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