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Differentiation from first principlesEdexcel A-Level Maths: Subtopic test

10 questions, 27 marks

Edexcel A-Level Maths

Differentiation from first principles

Total 27 marks

Name

Class

Date

  1. 1
    The point P(3,9)P(3,9) lies on the curve y=x2y=x^{2}. The point QQ on the same curve has xx-coordinate 3+h3+h, where h≠0h\neq0.
    (a)
    Find the gradient of the chord PQPQ.
    [1 mark]
    • A66
    • B3+h3+h
    • C6+h26+h^{2}
    • D6+h6+h
    (b)
    As hh tends to 00, the gradient of PQPQ tends to the gradient of the tangent at PP. What is this gradient?
    [1 mark]
    • A00
    • B33
    • C66
    • D99
    (c)
    Find the equation of the tangent to the curve at PP.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    Let f(x)=x3f(x)=x^{3}.
    (a)
    Which expression is equal to f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h} after simplifying?
    [1 mark]
    • A3x2+h23x^{2}+h^{2}
    • B3x2+3xh+h23x^{2}+3xh+h^{2}
    • C3x2+3xh+h33x^{2}+3xh+h^{3}
    • Dh2h^{2}
    (b)
    Hence, from first principles, what is f′(x)f'(x)?
    [1 mark]
    • A3x23x^{2}
    • B3x2+3xh+h23x^{2}+3xh+h^{2}
    • Cx2x^{2}
    • D3x3x
    (c)
    Given that f′(x)=3x2f'(x)=3x^{2}, find f′′(x)f''(x), and hence find the rate of change of the gradient of the curve y=f(x)y=f(x) when x=2x=2.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    A curve has equation y=x2y=x^{2}.
    (a)
    Prove, from first principles, that dydx=2x\frac{\mathrm{d}y}{\mathrm{d}x}=2x.
    [3 marks]
    (b)
    Find the coordinates of the point on the curve at which the gradient of the tangent is −6-6, and find the equation of the tangent at this point.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    A particle moves in a straight line. Its displacement from a fixed point OO after tt seconds is s=t3−6t2s=t^{3}-6t^{2} metres, for t≥0t\geq0. Its velocity is v=dsdtv=\frac{\mathrm{d}s}{\mathrm{d}t} and its acceleration is a=d2sdt2a=\frac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}.
    (a)
    (i) Show, from first principles, that the derivative of t3t^{3} with respect to tt is 3t23t^{2}.
    (ii) Given that the derivative of
    t2t^{2} is 2t2t, find expressions for vv and aa in terms of tt.
    [6 marks]
    (b)
    (i) Find the times at which the particle is instantaneously at rest.
    (ii) Find the acceleration when
    t=4t=4 and interpret its meaning.
    (iii) Find the gradient of the curve of
    ss against tt when t=1t=1 and interpret its meaning.
    [6 marks]

    Total for question 4: 12 marks

End of questions

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).