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How do you differentiate polynomials and use the derivative to find gradients, tangents and turning points?

Differentiate polynomials: find first and second derivatives, find gradients, tangents and normals, locate and classify stationary points, and solve optimisation problems.

A CCEA GCSE Further Mathematics answer on differentiation, covering the power rule, the second derivative, tangents and normals, finding and classifying stationary points, increasing and decreasing functions, and optimisation in the compulsory Pure unit.

Generated by Claude Opus 4.813 min answer

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  1. What this dot point is asking
  2. The power rule
  3. Tangents and normals
  4. Stationary points
  5. Classifying with the second derivative
  6. Increasing, decreasing and optimisation
  7. Why this matters

What this dot point is asking

Differentiation is the calculus of change, and CCEA GCSE Further Mathematics introduces it for polynomials. You must differentiate to find the gradient function, find first and second derivatives, use the gradient to write equations of tangents and normals, locate stationary points and classify them as maxima or minima, decide where a function is increasing or decreasing, and solve optimisation problems. This is a major Further Mathematics topic and one of the most heavily examined.

The power rule

Differentiation of a polynomial works term by term using the power rule. Before differentiating, write every term as a power of xx, including roots and reciprocals.

For example, y=4x3βˆ’2x2+7xβˆ’5y = 4x^3 - 2x^2 + 7x - 5 differentiates to dydx=12x2βˆ’4x+7\dfrac{dy}{dx} = 12x^2 - 4x + 7. The derivative is itself a function, so substituting an xx value gives the gradient at that point on the curve.

Tangents and normals

The gradient of the curve at a point equals the gradient of the tangent there. The normal is the line perpendicular to the tangent at that point, so its gradient is the negative reciprocal.

To find a tangent: differentiate, substitute the xx value to get the gradient, find the yy coordinate on the curve, then use yβˆ’y1=m(xβˆ’x1)y - y_1 = m(x - x_1). For the normal, replace the gradient with its negative reciprocal and use the same point.

Stationary points

A stationary point is where the curve momentarily levels off, so the gradient is zero. Setting the derivative to zero and solving gives their xx values; substituting back into the curve gives the coordinates.

Classifying with the second derivative

The second derivative is the derivative of the derivative, and its sign tells you the type of stationary point.

Increasing, decreasing and optimisation

A function is increasing where its derivative is positive and decreasing where its derivative is negative, so you can describe the behaviour of a curve by analysing the sign of dydx\dfrac{dy}{dx}. Optimisation problems use this: you write a quantity (such as area or volume) as a function of one variable, differentiate, set the derivative to zero to find the maximum or minimum, and check with the second derivative. These applied questions, for example maximising the volume of an open box, are a classic Further Mathematics challenge and carry several marks.

Why this matters

Differentiation underpins the whole calculus strand and reaches into the Mechanics unit, where differentiating displacement gives velocity and differentiating velocity gives acceleration. The same techniques find the turning points you also meet through completing the square, link to the tangent work of coordinate geometry, and pair with integration as the two halves of calculus. Optimisation in particular rewards a clear method: model, differentiate, solve, classify.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA Unit 1 (style)5 marksA curve has equation y=x3βˆ’3x2βˆ’9x+5y = x^3 - 3x^2 - 9x + 5. Find the coordinates of the stationary points.
Show worked answer β†’

Differentiate: dydx=3x2βˆ’6xβˆ’9\dfrac{dy}{dx} = 3x^2 - 6x - 9.

Set the derivative to zero for stationary points: 3x2βˆ’6xβˆ’9=03x^2 - 6x - 9 = 0, so x2βˆ’2xβˆ’3=0x^2 - 2x - 3 = 0, which factorises to (xβˆ’3)(x+1)=0(x - 3)(x + 1) = 0, giving x=3x = 3 or x=βˆ’1x = -1.

Find the yy values: at x=3x = 3, y=27βˆ’27βˆ’27+5=βˆ’22y = 27 - 27 - 27 + 5 = -22; at x=βˆ’1x = -1, y=βˆ’1βˆ’3+9+5=10y = -1 - 3 + 9 + 5 = 10.

The stationary points are (3,βˆ’22)(3, -22) and (βˆ’1,10)(-1, 10). Marks are for the derivative, solving it, and both coordinate pairs.

CCEA Unit 1 (style)4 marksFind the equation of the tangent to y=x2βˆ’4x+1y = x^2 - 4x + 1 at the point where x=3x = 3.
Show worked answer β†’

Find the yy coordinate: at x=3x = 3, y=9βˆ’12+1=βˆ’2y = 9 - 12 + 1 = -2, so the point is (3,βˆ’2)(3, -2).

Differentiate for the gradient: dydx=2xβˆ’4\dfrac{dy}{dx} = 2x - 4. At x=3x = 3 this is 2(3)βˆ’4=22(3) - 4 = 2.

Use yβˆ’y1=m(xβˆ’x1)y - y_1 = m(x - x_1): yβˆ’(βˆ’2)=2(xβˆ’3)y - (-2) = 2(x - 3), so y+2=2xβˆ’6y + 2 = 2x - 6, giving y=2xβˆ’8y = 2x - 8.

Marks are for the point, the derivative, the gradient value, and the tangent equation.

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