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Cambridge IGCSE0580

Differentiation

Mathematics 0580 Chapter Notes

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Differentiation
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In this chapter

  1. 1Estimating a gradient from a tangent
  2. 2Differentiating powers of x
  3. 3Gradient at a point
  4. 4Stationary points
  5. 5Maximum or minimum: the second derivative
  6. 6Maximum or minimum: other methods

1. Estimating a gradient from a tangent

The gradient of a curve at a point is the gradient of the tangent to the curve at that point. A tangent touches the curve at that point.

To estimate the gradient, draw the tangent. Then take two points on the tangent and find the change in y divided by the change in x.

gradient = (change in y)/(change in x)

Examiner insight

“It was quite common to see either a chord or a line that did not quite touch the curve at x = 2.” “After drawing a good tangent, some candidates then chose points on the curve to find a gradient, rather than finding the gradient of the tangent.”

Worked example 12 marks

A tangent is drawn to a curve at the point where x = 2. The tangent passes through (0, -1) and (3, 11). Estimate the gradient of the curve at x = 2.

  1. 1

    gradient of curve = gradient of tangent

  2. 2

    (11 - (-1))/(3 - 0) = 12/3 [1]

  3. 3

    = 4 [1]

2. Differentiating powers of x

The derivative dy/dx gives the gradient of the curve at any point. To differentiate axⁿ, multiply by the power and reduce the power by one.

A function of up to three terms is differentiated term by term. The derivative of ax is a. The derivative of a constant is 0.

y = axⁿ ⇒ dy/dx = anxⁿ⁻¹

Examiner insight

“Some candidates made errors when differentiating, such as incorrect terms 2x2, –5 instead of –5x and the inclusion of the constant term 3.” “The most common errors were forgetting to multiply the coefficient by the index, hence giving −2x rather than −4x and to leave in the constant of 5 or change this to 1 in the derivative.”

Worked example 13 marks

A curve has equation y = x³ - 5/2x² + 3. Find the gradient of the curve at x = 1.5.

  1. 1

    dy/dx = 3x² - 5x [2]

  2. 2

    at x = 1.5: 3(1.5)² - 5(1.5) = 6.75 - 7.5 = -0.75 [1]

3. Gradient at a point

The derivative gives the gradient of the curve, and so of the tangent, at any point. Substitute the x-value into dy/dx.

Substituting into y does not give the gradient. If a question gives the gradient, set the derivative equal to that value.

gradient of tangent at x = a is dy/dx with x = a

Examiner insight

“It was clear, however, that many did not know that the derivative of the function gave the gradient, and these started by substituting x = 2 and y = 6 into the given cubic function.” “Some recognised the need to find the gradient of the tangent but not all linked this with using the derivative, often finding the gradient of a chord instead.”

Worked example 13 marks

The equation of a curve is y = 2x² - 11x + 12. Find the gradient of the tangent to the curve at x = 2.

  1. 1

    dy/dx = 4x - 11 [2]

  2. 2

    at x = 2: 4(2) - 11 = -3 [1]

4. Stationary points

A stationary point (turning point) is where the gradient is zero. Find dy/dx and set it equal to 0. Solve for x.

Then substitute each x-value into the original equation for y to find the coordinates.

dy/dx = 0

Examiner insight

“Some candidates went on to equate their first derivative to 0 but of these, some had difficulty factorising to find the solutions for x and others omitted the final step of substituting their x values into the given equation to find the y values.” “Some weaker candidates equated the equation of the graph to 0 rather than the derivative.”

Worked example 15 marks

Find the coordinates of the turning points of the curve y = 8x³ - 6x + 5. You must show all your working.

  1. 1

    dy/dx = 24x² - 6 [2]

  2. 2

    set dy/dx = 0: 24x² = 6, so x = ±1/2 [1]

  3. 3

    x = 1/2: y = 8 × 1/8 - 3 + 5 = 3, so (1/2, 3) [1]

  4. 4

    x = -1/2: y = -1 + 3 + 5 = 7, so (-1/2, 7) [1]

5. Maximum or minimum: the second derivative

Differentiate again to get the second derivative d²y/dx². Substitute the x-value of each stationary point into it.

If the value is positive the point is a minimum. If it is negative the point is a maximum. Write the value and the conclusion.

The second derivative is not set equal to 0. Only dy/dx is, to find the points.

d²y/dx² > 0 ⇒ minimum

d²y/dx² < 0 ⇒ maximum

Examiner insight

“As candidates were asked to show how they decided it was important to show the substitution into the second derivative, its evaluation and a comment on the value.” “Some that did make the comment then spoiled their answer by statements referring to the x-values rather than the values of the second derivative.”

Worked example 13 marks

Determine whether each turning point of the curve y = 8x³ - 6x + 5 is a maximum or a minimum. Show how you decide.

  1. 1

    d²y/dx² = 48x [1]

  2. 2

    x = 1/2: 48 × 1/2 = 24 > 0, so (1/2, 3) is a minimum [1]

  3. 3

    x = -1/2: 48 × (-1/2) = -24 < 0, so (-1/2, 7) is a maximum [1]

6. Maximum or minimum: other methods

A second method is an accurate sketch. A cubic with a positive x³ term rises, then falls, then rises. Its first turning point is a maximum and its second is a minimum.

A third method is to inspect the gradient either side of the point. A gradient that changes from positive to negative gives a maximum. Negative to positive gives a minimum.

Examiner insight

“The alternative efficient method to gain all marks is to draw a reasonable sketch of a positive cubic and write the co-ordinates of the stationary points in full with the conclusion of maximum or minimum, but this was not seen very often.” “It was very rare indeed to see any attempts at either finding the values of y on both sides of each of the turning points or calculating the gradients on either side of the turning points.”

Worked example 13 marks

The graph of y = x³ - 6x² + 7 has turning points at (0, 7) and (4, -25). By drawing a suitable sketch, determine whether each turning point is a maximum or a minimum.

  1. 1

    sketch a positive cubic through both turning points [1]

  2. 2

    (0, 7) is the first turning point: a maximum [1]

  3. 3

    (4, -25) is the second turning point: a minimum [1]

Recap

  • Extended: the gradient of a curve at a point is the gradient of the tangent there.
  • Extended: to differentiate axⁿ, multiply by n and reduce the power by one. A constant differentiates to 0.
  • Extended: gradient at a point = dy/dx with the x-value substituted.
  • Extended: stationary points: set dy/dx = 0, solve, then find y from the original equation.
  • Extended: d²y/dx² > 0 gives a minimum and d²y/dx² < 0 gives a maximum.
  • Extended: a sketch of a positive cubic or the gradient either side also decides maximum or minimum.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Do not use a calculator. Write down dy/dx when y = 4x⁵.1 mark
  2. Do not use a calculator. Differentiate y = 3x² - 7x + 4.2 marks
  3. Do not use a calculator. Find dy/dx when y = 1/2x⁴ - 2/3x³ + 9.2 marks
  4. A tangent is drawn to a curve at the point where x = 4. The tangent passes through (1, 7) and (5, 3). Estimate the gradient of the curve at x = 4.2 marks
  5. Do not use a calculator. Find the gradient of the curve y = x³ - 4x at x = -2.3 marks
  6. Do not use a calculator. The equation of a curve is y = x³ - 3x² + 2x. Find the values of x at which the gradient of the curve is 2.4 marks
  7. Do not use a calculator. Find the coordinates of the turning points of the curve y = 2x³ - 3x² - 12x + 1.5 marks
  8. Do not use a calculator. The gradient of the curve y = 2x³ - 9x² + 12x is zero at x = 1 and at x = 2. Use the second derivative to decide whether each is a maximum or a minimum.4 marks
  9. Do not use a calculator. The curve y = 5 + kx² - x³, where k is a constant, has a turning point where x = 2. Find the value of k. Find the coordinates of both turning points and determine whether each is a maximum or a minimum.7 marks
  10. Do not use a calculator. An open box has height x cm and volume y cm³, where y = 4x³ - 48x² + 144x and 0 < x < 6. Find the height that gives the greatest volume and show that it gives a maximum. Find the greatest volume.7 marks

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