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Differentiation

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Lesson notes

Differentiation Basics

  • Differentiation changes a curve equation y = \ldots into a gradient function \frac{dy}{dx} = \ldots.
  • To differentiate y = xn, bring down the power and reduce it by one: \frac{dy}{dx} = n xn-1.
  • For y = kxn, multiply by the coefficient: \frac{dy}{dx} = kn xn-1.
  • Special cases: y = kx gives \frac{dy}{dx} = k; y = c (constant) gives \frac{dy}{dx} = 0.
  • Differentiate each term separately when a curve has multiple terms.

Finding the Gradient at a Point

  • To find the gradient at a point, substitute the x-coordinate into \frac{dy}{dx}.
  • If given a gradient, set \frac{dy}{dx} equal to that value and solve for x.
  • The y-coordinate is not needed to find the gradient.

Gradient at a point via the tangent

y = x³ − 2x² + 6O−224681012−2−11234xytangent at x= −1,gradient 7

Stationary Points & Turning Points

  • A stationary point occurs where the gradient is zero: \frac{dy}{dx} = 0.
  • Turning points are stationary points where the curve changes direction (maxima or minima).
  • To find coordinates: (1) differentiate, (2) set \frac{dy}{dx} = 0 and solve for x, (3) substitute x into original equation to get y.

Turning points on a cubic curve

y = x³ − 3xO−4−3−2−11234−3−2−1123xyMax (−1, 2)Min (1, −2)

Classifying Stationary Points Using Graphs

  • A positive quadratic (x2 positive) has a minimum; a negative quadratic has a maximum.
  • A positive cubic has a maximum on the left and a minimum on the right.
  • A negative cubic has a minimum on the left and a maximum on the right.

Classifying turning points on a cubic

Positive vs negative cubicO−4−3−2−11234−3−2−1123xyMax (−1, 2)Min (−1, −2)Max (1, 2)negativepositiveMin (1, −2)

Classifying Using the First Derivative

  • Examine the sign of \frac{dy}{dx} just before and after the stationary point.
  • If gradient changes from positive to zero to negative → maximum.
  • If gradient changes from negative to zero to positive → minimum.

Classifying Using the Second Derivative

  • The second derivative \frac{d2y}{dx2} is the derivative of \frac{dy}{dx}.
  • Substitute the x-coordinate of the stationary point into \frac{d2y}{dx2}.
  • If \frac{d2y}{dx2} < 0maximum; if \frac{d2y}{dx2} > 0minimum; if zero, test fails.

Problem Solving with Differentiation (Optimisation)

  • Use differentiation to find maximum or minimum values of quantities (e.g., area, volume).
  • Form an equation for the quantity in terms of one variable, then differentiate and set \frac{dy}{dx} = 0.
  • Solve for the variable, then substitute back to find the optimum value.
  • Check if it is a max or min using second derivative or graph shape.

A cuboid with variable dimensions

CuboidLengthHeightWidth

Slides

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Practice questions

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  1. 1.What is the derivative of y = x5?

    Easy
    • A5x4
    • B5x5
    • Cx4
    • D4x5
  2. 2.What is the derivative of y = 2x3?

    Easy
    • A6x2
    • B5x2
    • C2x2
    • D6x3
  3. 3.What is the derivative of y = 4?

    Easy
    • A0
    • B4
    • C1
    • D4x
  4. 4.What is the derivative of y = 6 + 4x - x2?

    Easy
    • A4 - 2x
    • B4 + 2x
    • C6 + 4x - 2x
    • D4x - 2x
  5. 5.Find the gradient of y = 24 + 5x - x2 at x = -1.5.

    Medium
    • A8
    • B2
    • C-8
    • D-2
  6. 6.Given y = 2xk + u x7 and dy/dx = 18 xk-1 + 21 x6, find k and u.

    Medium
    • Ak=9, u=3
    • Bk=9, u=7
    • Ck=18, u=21
    • Dk=2, u=3
  7. 7.Find the x-coordinate of the turning point of y = 6 + 4x - x2.

    Medium
    • A2
    • B-2
    • C4
    • D6
  8. 8.Find the coordinates of the turning point of y = 2x2 + 8x - 9.

    Medium
    • A(-2, -17)
    • B(2, 15)
    • C(-2, 1)
    • D(2, -17)

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