Mathematics · OCR H240 · Topic 1.07

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
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What the specification says

Every sub-topic OCR 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. 1.07a-hGradients and differentiation from first principlesY13
  2. 1.07i-lDifferentiation of standard functionsY13
  3. 1.07m-pTangents, normals, stationary points, increasing and decreasing functionsY13
  4. 1.07q-rTechniques of differentiationY13
  5. 1.07sParametric and implicit differentiationY13
  6. 1.07tConstructing 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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