Mathematics · Pearson Edexcel 9MA0 · Topic 7

Differentiation

Finding gradients and rates of change: the rules for differentiating, tangents, stationary points, the product, quotient and chain rules, and implicit and parametric curves.

  • 6 sub-topics
  • Taught in Year 13
  • 118 practice questions
  • 6 animated lessons
Practice questions (118) Lessons Worksheets All Edexcel Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 7.1Gradients, first principles and second derivativesY13
  2. 7.2Differentiating standard functionsY13
  3. 7.3Tangents, normals and stationary pointsY13
  4. 7.4The product, quotient and chain rulesY13
  5. 7.5Implicit and parametric differentiationY13
  6. 7.6Construct simple differential equationsY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 118 questions in this topic, free, no account needed.

Differentiation

Easy

Gradient of a curve · Year 12 · 3 marks

A curve has equation y=2x3−5x2+4y = 2x^3 - 5x^2 + 4.
  1. Find the gradient of the curve at the point where x=2x = 2.
    (3)
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  1. Differentiate each term: multiply by the power, then reduce the power by 1. The constant disappears.dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}=6x2−10x{}= 6x^2 - 10x
  2. The gradient at a point is the derivative evaluated there.At x=2x = 2: 24−20=424 - 20 = 4

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Differentiation

Medium

Stationary points · Year 12 · 6 marks

A curve has equation y=x3−6x2+9x+1y = x^3 - 6x^2 + 9x + 1.
  1. Find the coordinates of the stationary points of the curve and determine their nature.
    (6)
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  1. Stationary points are where the gradient is zero, so differentiate first.dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}=3x2−12x+9{}= 3x^2 - 12x + 9
  2. Set the derivative to zero and solve.3(x−1)(x−3)3(x - 1)(x - 3)=0{}= 0⇒x{}\Rightarrow x=1, x{}= 1,\ x=3{}= 3
  3. Substitute each xx into the original equation to get yy.(1,5)(1, 5) and (3,1)(3, 1)
  4. The second derivative tells you the shape at each point.d2ydx2\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=6x−12{}= 6x - 12
  5. Negative second derivative means the curve is bending down.At x=1x = 1: −6<0-6 < 0, so (1,5)(1, 5) is a maximum.
  6. Positive second derivative means the curve is bending up.At x=3x = 3: 6>06 > 0, so (3,1)(3, 1) is a minimum.

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Differentiation

Hard

Making the biggest box · Year 12 · 7 marks

An open box is made from a 24 cm square of card by cutting a square of side xx cm from each corner and folding up the sides.
  1. (a)
    Show that the volume is V=4x3−96x2+576xV = 4x^3 - 96x^2 + 576x.
    (2)
  2. (b)
    Find the maximum volume, showing that it is a maximum.
    (5)
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Part (a)2 marks

  1. The base is 24−2x24 - 2x each way and the height is xx.V=x(24−2x)2V = x(24 - 2x)^2
  2. =x(576−96x+4x2)= x(576 - 96x + 4x^2)=4x3−96x2+576x{}= 4x^3 - 96x^2 + 576x

Part (b)5 marks

  1. dVdx\dfrac{\mathrm{d}V}{\mathrm{d}x}=12x2−192x+576{}= 12x^2 - 192x + 576=12(x−4)(x−12){}= 12(x - 4)(x - 12)
  2. When x=12x = 12 the base has width 0.x=4x = 4 (x=12x = 12 gives no box)
  3. Negative second derivative means a maximum.d2Vdx2\dfrac{\mathrm{d}^2V}{\mathrm{d}x^2}=24x−192{}= 24x - 192=−96<0{}= -96 < 0, so a maximum
  4. V=4(16)2=1024V = 4(16)^2 = 1024 cm³

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