Differentiation: Gradients and Turning Points
Differentiation finds the gradient function of a curve. For a term of the form ax^n the derivative is anx^(n-1), so differentiating each term of a polynomial gives dy/dx, and substituting an x value into dy/dx gives the gradient of the curve at that point. Turning points are where the gradient is zero, so solving dy/dx = 0 finds them. Their nature is settled either by the second derivative (negative for a maximum, positive for a minimum) or by testing the sign of dy/dx just either side. This is IGCSE content that does not appear on the GCSE specification at all, and it also underlies tangents, normals and rates of change.
Before you start
No specific prerequisites - this is a good place to start.
Method
- Write the expression as a sum of terms in the form ax^n, converting roots and fractions first, since 1/x must be written as x^(-1) and the square root of x as x^(1/2) before the rule applies.
- Differentiate term by term with the rule: the derivative of ax^n is anx^(n-1). The derivative of a constant is zero, because a constant graph has no gradient.
- For the gradient at a point, substitute the x value into dy/dx. For the equation of the tangent, use that gradient with the point in y - y1 = m(x - x1); for the normal, use the negative reciprocal of the gradient.
- For turning points, set dy/dx = 0 and solve, usually by factorising the quadratic you get.
- Find the y coordinate of each turning point by substituting the x value back into the ORIGINAL equation, not into dy/dx, which is the most common lost mark.
- Determine the nature: differentiate again to get the second derivative and substitute each x value. A negative second derivative means a maximum, a positive one a minimum. If it is zero, test the sign of dy/dx just below and just above instead.
Worked example
A curve has equation y = x^3 - 6x^2 + 9x. Find the coordinates of the two turning points and determine the nature of each.
- Differentiate term by term: dy/dx = 3x^2 - 12x + 9.
- Set the gradient to zero: 3x^2 - 12x + 9 = 0. Divide through by 3 to get x^2 - 4x + 3 = 0, then factorise to (x - 1)(x - 3) = 0, so x = 1 or x = 3.
- Substitute into the original equation for the y values: when x = 1, y = 1 - 6 + 9 = 4, and when x = 3, y = 27 - 54 + 27 = 0.
- Differentiate again for the nature: the second derivative is 6x - 12.
- At x = 1 the second derivative is 6 - 12 = -6, which is negative, so (1, 4) is a maximum. At x = 3 it is 18 - 12 = 6, which is positive, so (3, 0) is a minimum.
- Final answer: a maximum at (1, 4) and a minimum at (3, 0).
Practice questions
Try each question, then tap to reveal the answer.
Q1Differentiate y = 4x^3 - 7x + 2.Show answer
Answer: dy/dx = 12x^2 - 7.
Q2Differentiate y = 5x^2 + 3/x, writing the second term as a power of x first.Show answer
Answer: Write it as 5x^2 + 3x^(-1), so dy/dx = 10x - 3x^(-2), which is 10x - 3/x^2.
Q3Find the gradient of the curve y = x^2 - 4x at the point where x = 5.Show answer
Answer: dy/dx = 2x - 4, so at x = 5 the gradient is 10 - 4 = 6.
Q4Why must the y coordinate of a turning point be found from the original equation rather than from dy/dx?Show answer
Answer: dy/dx gives the gradient, not the height of the curve. At a turning point the gradient is zero, so substituting into dy/dx would always give 0 rather than the y value.
Q5The second derivative of a curve at a stationary point is 8. What is the nature of that point?Show answer
Answer: It is a minimum, because the second derivative is positive.
Q6Find the coordinates of the stationary point of y = x^2 - 6x + 5.Show answer
Answer: dy/dx = 2x - 6 = 0 gives x = 3; then y = 9 - 18 + 5 = -4, so the stationary point is (3, -4).
Q7Find the equation of the tangent to y = x^2 at the point (3, 9).Show answer
Answer: dy/dx = 2x, so the gradient at x = 3 is 6. Using y - 9 = 6(x - 3) gives y = 6x - 9.
Exam-style questions
Written in the style of a IGCSE Maths exam paper, with a full mark scheme.
A curve has equation y = 2x^3 - 3x^2 - 12x + 5. Find the x coordinates of the stationary points and determine the nature of each.
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Find the equation of the normal to the curve y = x^2 - 3x + 4 at the point where x = 2. Give your answer in the form y = mx + c.
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Free printable worksheet
Want more practice on paper? Download the differentiation: gradients and turning points worksheet pack - 9 pages of exam-style questions with a full mark scheme. One email opens every download in this browser for 14 days - no account, no card. Print it for personal and classroom use.
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