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5.3 Differentiating polynomialsIB Maths: Applications and Interpretation SL: Revision notes

Section 1

The derivative as a gradient function

The derivative of a function gives the gradient of its graph at every point. It is written f′(x)f'(x) or dydx\frac{dy}{dx} when y=f(x)y=f(x). To find the gradient at a particular point, differentiate first and then substitute the xx-value into the derivative. The derivative is also a rate of change: if C(x)C(x) is a cost in AED and xx is the number of items, then C′(x)C'(x) is the rate of change of cost in AED per item.

Key termsderivativegradient functionrate of change
Common mistake

Substituting the xx-value into f(x)f(x) instead of f′(x)f'(x). That gives the height of the curve, not its gradient.

Section 2

The power rule

If f(x)=axnf(x)=ax^n, where nn is an integer, then f′(x)=anxn−1.f'(x)=anx^{n-1}. Multiply by the power, then reduce the power by one. Special cases: f(x)=kxf(x)=kx gives f′(x)=kf'(x)=k, and a constant f(x)=kf(x)=k gives f′(x)=0f'(x)=0 (a horizontal line has gradient zero). Examples: ddx(5x4)=20x3\frac{d}{dx}(5x^4)=20x^3; ddx(7x)=7\frac{d}{dx}(7x)=7; ddx(9)=0\frac{d}{dx}(9)=0.

Key termspower rule
Common mistake

Differentiating a constant term to get the constant back. It should disappear.

Exam tip

Say it as you go: 'times the power, then power minus one'.

Section 3

Sums of terms and negative powers

Differentiate a sum term by term: for f(x)=axn+bxn−1+…f(x)=ax^n+bx^{n-1}+\ldots, differentiate each term separately and add the results. The power rule also works when nn is negative. First rewrite 1xk\frac{1}{x^k} as x−kx^{-k}. For f(x)=3x2=3x−2f(x)=\frac{3}{x^2}=3x^{-2}, f′(x)=−6x−3=−6x3f'(x)=-6x^{-3}=-\frac{6}{x^3}. Subtracting one from a negative power makes it more negative. If the function is a product or a fraction, expand or divide first so that every term has the form axnax^n. For f(x)=(2x+1)(x−3)=2x2−5x−3f(x)=(2x+1)(x-3)=2x^2-5x-3, f′(x)=4x−5f'(x)=4x-5. For f(x)=x2+3x=x+3x−1f(x)=\frac{x^2+3}{x}=x+3x^{-1}, f′(x)=1−3x−2f'(x)=1-3x^{-2}.

Key termsterm by termnegative power
Common mistake

Writing ddx(x−2)=−2x−1\frac{d}{dx}(x^{-2})=-2x^{-1}. The power must go down to −3-3, not up to −1-1.

Exam tip

Rewrite as powers of xx before differentiating, then convert back at the end if the question wants fractions.

Section 4

Using the derivative

Worked example: f(x)=2x3−5x2+x−4f(x)=2x^3-5x^2+x-4. Then f′(x)=6x2−10x+1f'(x)=6x^2-10x+1.

  • Gradient at x=2x=2: f′(2)=24−20+1=5f'(2)=24-20+1=5.
  • Where is the gradient 11? Solve 6x2−10x+1=16x^2-10x+1=1, so 2x(3x−5)=02x(3x-5)=0 and x=0x=0 or x=53x=\frac{5}{3}.
  • Increasing or decreasing at a point: the sign of f′f' at that point (positive means increasing). Always check you have found every solution: dividing both sides by xx would lose the solution x=0x=0. Your GDC can solve f′(x)=kf'(x)=k directly.
Exam tip

Set f′(x)=kf'(x)=k and bring everything to one side before factorising or using the GDC.

Section 5

Setting out and common slips

Write the derivative on its own line with correct notation, for example f′(x)=12x2−12x+5f'(x)=12x^2-12x+5 or dydx=…\frac{dy}{dx}=\ldots; do not write f(x)=12x2…f(x)=12x^2\ldots. Final answers in context need units: a rate of change of cost in AED per item, or of surface area in cm2^2 per cm. Give exact values or three significant figures. Check your work: each term's power should drop by one; a x1x^1 term becomes a constant; the constant term vanishes.

Common mistake

Mixing up f(a)f(a) and f′(a)f'(a): the first is the height of the curve at x=ax=a, the second is its gradient.

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Exam questions on 5.3 Differentiating polynomials

  1. A function is defined by f(x)=4x3−6x2+5x−9f(x)=4x^3-6x^2+5x-9.
    Find the values of xx at which the gradient of the graph of ff is 55.2 marks
  2. A function is defined by g(x)=3x2+4x−2x2g(x)=3x^2+\frac{4}{x}-\frac{2}{x^2}, for x≠0x\neq0.
    Find g′(−1)g'(-1) and state whether gg is increasing or decreasing at x=−1x=-1.2 marks
  3. The cost, CC AED, of producing xx items in a day is modelled by C(x)=0.5x3−6x2+40x+200C(x)=0.5x^3-6x^2+40x+200, for 0<x≤200<x\le20. The rate of change of cost, in AED per item, is given by C′(x)C'(x).
    (i) Find C′(x)C'(x). (ii) Find the rate of change of cost when 1010 items are produced.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).