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A-Level Differentiation

What the A-Level syllabus expects for Differentiation, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to find, determine, solve, evaluate. About 8% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (5 marks)

The curve C has equation y = (x^2 + 12)/sqrt(x) for x > 0. By differentiating, demonstrate that C possesses a stationary point at (2, 16/sqrt(2)), and establish whether it is a maximum, a minimum or a point of inflexion.

Show the worked answer

y = (x² + 12)/x^(1/2) = x^(3/2) + 12 x^(-1/2). dy/dx = (3/2)x^(1/2) - 6 x^(-3/2). Set = 0: (3/2)x^(1/2) = 6 x^(-3/2) => (3/2)x² = 6 => x² = 4 => x = 2 (x>0). Then y = (4+12)/sqrt2 = 16/sqrt2, so the stationary point is (2, 16/sqrt2). d^2y/dx² = (3/4)x^(-1/2) + 9 x^(-5/2). At x=2 both terms are positive, so d^2y/dx² > 0 => minimum.

Example 2 (3 marks)

The watch retailer also wishes to model its yearly promotion spending, C thousand dollars per year. For t years with t >= 0, the model is C = t^3 - 12t^2 + (k + 36)t, in which k is a positive constant. Determine the set of k-values for which C increases with t.

Show the worked answer

C = t³ - 12t² + (k + 36)t. Then dC/dt = 3t² - 24t + (k + 36). C increases with t on t >= 0 provided dC/dt >= 0 for all t >= 0. The quadratic 3t² - 24t + (k + 36) has its minimum at t = -(-24)/(2*3) = 4, which lies in the domain t >= 0. Minimum value: 3(4)² - 24(4) + (k + 36) = 48 - 96 + k + 36 = k - 12. Require k - 12 >= 0, i.e. k >= 12. (Given k > 0, this is the binding condition.)

Example 3 (3 marks)

A bakery's monthly revenue R (in thousands of dollars) is described by the model R = 3000 - 6000/(3 + 0.5t), where t denotes the number of years elapsed since 2015. Using differentiation, determine the value of dR/dt when t = 2, and interpret what this quantity tells you about the bakery's revenue in the given situation.

Show the worked answer

R = 3000 - 6000(3 + 0.5t)⁻¹. Differentiate: dR/dt = -6000 * (-1)(3 + 0.5t)⁻² * 0.5 = 3000/(3 + 0.5t)². At t = 2: 3 + 0.5(2) = 4, so dR/dt = 3000/4² = 3000/16 = 187.5. Since R is in thousands of dollars, this is +187.5 thousand dollars per year, i.e. $187,500 per year. Positive value means revenue is increasing.

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More A-Level H2 Maths topics

Functions · Graphs and their transformations · Equations and inequalities · Sequences and series · Vectors in two and three dimensions: basic properties · Scalar and vector products · all of A-Level H2 Maths