Differentiation in IGCSE Maths
The power rule, dy/dx notation, finding the gradient of a curve at a point, and using stationary points to locate maximum and minimum values. This is Extended-tier content, with clear graphs, worked examples and an exam-style question.
0580 · E2.12 ExtendedEstimating a gradient with a tangent
A straight line has the same gradient everywhere, but the gradient of a curve changes from point to point. To estimate the gradient at a particular point, draw a tangent, a straight line that just touches the curve at that point, and find the gradient of the tangent.
Here the tangent touches \(y = x^2\) at \((2, 4)\). Its gradient is the rise divided by the run, which works out as \(4\). Differentiation gives us that same answer exactly, by algebra, with no drawing needed.
The tangent should touch the curve at one point only and lean the same way as the curve there. Draw it long, then use two points far apart to work out the gradient, since a short line gives a much less accurate answer.
The power rule
Differentiating an expression gives the derived function, or derivative, written \(\dfrac{dy}{dx}\). It is a formula for the gradient of the curve at any value of \(x\). The rule is:
\[y = ax^n \quad \Rightarrow \quad \frac{dy}{dx} = anx^{n-1}\]In words: multiply by the power, then subtract \(1\) from the power.
Differentiate \(y = x^5\) and \(y = 4x^3\).
For \(y = x^5\), multiply by \(5\) and reduce the power to \(4\):
\[\frac{dy}{dx} = 5x^4\]For \(y = 4x^3\), multiply \(4\) by \(3\) and reduce the power to \(2\):
\[\frac{dy}{dx} = 12x^2\]A term in \(x\) on its own differentiates to its coefficient, because \(x = x^1\). So \(y = 7x\) gives \(\dfrac{dy}{dx} = 7\).
A constant differentiates to \(0\), because a horizontal line has zero gradient. So \(y = 9\) gives \(\dfrac{dy}{dx} = 0\).
Differentiating sums of terms
When an expression has several terms, differentiate each term separately and keep the signs. At IGCSE you will meet sums of up to three of these terms, with whole-number powers.
Differentiate \(y = 3x^2 + 5x - 2\).
Take the terms one at a time. \(3x^2\) gives \(6x\), \(5x\) gives \(5\), and the constant \(-2\) gives \(0\):
\[\frac{dy}{dx} = 6x + 5\]The gradient at a point
To find the gradient of a curve at a particular point, differentiate first, then substitute the \(x\)-coordinate of that point into \(\dfrac{dy}{dx}\).
Find the gradient of the curve \(y = x^2 - 4x + 1\) at the point where \(x = 3\).
Differentiate:
\[\frac{dy}{dx} = 2x - 4\]Substitute \(x = 3\):
\[\frac{dy}{dx} = 2(3) - 4 = 2\]The gradient at that point is \(2\).
A positive gradient means the curve is going uphill at that point, a negative gradient means downhill, and a gradient of zero means the curve is momentarily flat.
Stationary points
A stationary point, also called a turning point, is where the curve is flat, so the gradient is zero. To find stationary points, set the derivative equal to zero and solve:
\[\frac{dy}{dx} = 0\]That gives the \(x\)-coordinates. Substitute each one back into the original equation to get the matching \(y\)-coordinates.
Find the coordinates of the stationary points on the curve \(y = x^3 - 3x^2 + 4\).
Differentiate and set equal to zero:
\[\frac{dy}{dx} = 3x^2 - 6x = 0\]Factorise and solve:
\[3x(x - 2) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 2\]Substitute back into the original equation:
\[x = 0: \ y = 0 - 0 + 4 = 4\] \[x = 2: \ y = 8 - 12 + 4 = 0\]The stationary points are \((0, 4)\) and \((2, 0)\).
Maximum or minimum?
Once you have a stationary point you must say whether it is a maximum or a minimum. Any of these three methods is accepted:
1. Sketch the curve. A peak is a maximum, a dip is a minimum.
2. Check the gradient either side. Substitute an \(x\)-value just below and just above the turning point into \(\dfrac{dy}{dx}\). Positive then negative means a maximum; negative then positive means a minimum.
3. Use the second derivative. Differentiate a second time to get \(\dfrac{d^2y}{dx^2}\). If it is negative the point is a maximum; if it is positive the point is a minimum.
Determine the nature of the stationary points of \(y = x^3 - 3x^2 + 4\), found above at \((0, 4)\) and \((2, 0)\).
Differentiate \(\dfrac{dy}{dx} = 3x^2 - 6x\) again:
\[\frac{d^2y}{dx^2} = 6x - 6\]At \(x = 0\): \(6(0) - 6 = -6\), which is negative, so \((0, 4)\) is a maximum.
At \(x = 2\): \(6(2) - 6 = 6\), which is positive, so \((2, 0)\) is a minimum.
This matches the graph above.
Negative second derivative gives a maximum, positive gives a minimum. Many students expect the opposite, so it is worth learning deliberately. A quick sketch is a good way to check your answer.
Exam-style question
A curve has equation \(y = 2x^3 - 3x^2 - 12x + 5\).
(a) Find \(\dfrac{dy}{dx}\). [2]
(b) Find the coordinates of the two stationary points. [4]
(c) Determine whether each stationary point is a maximum or a minimum. [2]
Show solution (a)
Differentiate each term in turn:
\[\frac{dy}{dx} = 6x^2 - 6x - 12\]Show solution (b)
Set the derivative to zero and divide through by \(6\):
\[6x^2 - 6x - 12 = 0 \quad \Rightarrow \quad x^2 - x - 2 = 0\]Factorise:
\[(x - 2)(x + 1) = 0 \quad \Rightarrow \quad x = 2 \text{ or } x = -1\]Substitute back into the original equation:
\[x = -1: \ y = -2 - 3 + 12 + 5 = 12\] \[x = 2: \ y = 16 - 12 - 24 + 5 = -15\]The stationary points are \((-1, 12)\) and \((2, -15)\).
Show solution (c)
Differentiate again:
\[\frac{d^2y}{dx^2} = 12x - 6\]At \(x = -1\): \(12(-1) - 6 = -18\), negative, so \((-1, 12)\) is a maximum.
At \(x = 2\): \(12(2) - 6 = 18\), positive, so \((2, -15)\) is a minimum.
This is the standard Paper 4 differentiation question and it follows the same three steps almost every time: differentiate, set to zero and solve, then test the nature of each point. The marks in part (b) are lost most often by finding the \(x\)-values and forgetting to substitute back for \(y\), so always finish with full coordinates.
Common mistakes
Setting \(\dfrac{dy}{dx} = 0\) gives only the \(x\)-values. Substitute each one back into the original equation to get \(y\), and give the answer as coordinates.
The \(y\)-coordinate comes from the original curve, not from the derivative. Putting the \(x\)-value into \(\dfrac{dy}{dx}\) gives the gradient, which is zero at a turning point.
A constant term differentiates to \(0\), and a term such as \(5x\) differentiates to \(5\). Leaving the constant unchanged is a very common slip.
A negative second derivative means a maximum and a positive one means a minimum. If you are unsure, check the gradient either side of the point or sketch the curve.
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