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The derivative as a gradient and rate of changeEdexcel International A Level Maths: Revision notes

Section 1

The gradient of a curve at a point

The gradient of a curve at a point is defined as the gradient of the tangent to the curve at that point: the straight line that just touches the curve there and has the same direction as the curve. For a curve y=f(x)y=f(x) the gradient changes from point to point, so we describe it with a gradient function, the derivative dydx\frac{\mathrm{d}y}{\mathrm{d}x}, also written f′(x)f'(x). Its value at x=ax=a is f′(a)f'(a), the gradient of the tangent at x=ax=a. If f′(a)>0f'(a)>0 the curve is rising as xx increases, if f′(a)<0f'(a)<0 it is falling, and if f′(a)=0f'(a)=0 the tangent is horizontal.

Key termstangentgradient functionderivative
Common mistake

Confusing the value of yy at a point with the gradient there. f(2)f(2) is a height; f′(2)f'(2) is a gradient.

Section 2

The gradient as a limit

A chord joins two points on a curve. For the points with xx-coordinates xx and x+hx+h its gradient is f(x+h)−f(x)h.\frac{f(x+h)-f(x)}{h}. This is only an average gradient. As hh gets smaller the second point moves towards the first and the chord approaches the tangent. The derivative is defined as the limit f′(x)=lim⁡h→0f(x+h)−f(x)h.f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}. Using this definition is called differentiation from first principles. Example: f(x)=x2f(x)=x^2. (x+h)2−x2h=2xh+h2h=2x+h\frac{(x+h)^2-x^2}{h}=\frac{2xh+h^2}{h}=2x+h. As h→0h\to0, f′(x)=2xf'(x)=2x.

Key termschordlimitfirst principles
Exam tip

Expand f(x+h)f(x+h) fully, cancel the terms with no hh, then divide every remaining term by hh before letting h→0h\to0.

Common mistake

Letting h=0h=0 before dividing by hh. That gives 00\frac00; divide first.

Section 3

Notation and rate of change

The derivative of yy with respect to xx is written dydx\frac{\mathrm{d}y}{\mathrm{d}x} or, if y=f(x)y=f(x), f′(x)f'(x). It measures the rate of change of yy with respect to xx: how fast yy changes per unit increase in xx. Its units are the units of yy divided by the units of xx. Example: the volume VV cm3^3 of water at time tt minutes is V=40+6t−0.5t2V=40+6t-0.5t^2. Then dVdt=6−t\frac{\mathrm{d}V}{\mathrm{d}t}=6-t in cm3^3 min−1^{-1}. At t=4t=4 the volume is increasing at 22 cm3^3 min−1^{-1}, and at t=6t=6 the rate is zero. A negative rate means the quantity is decreasing.

Key termsrate of change
Exam tip

State the units of a rate of change as 'units of yy per unit of xx', for example cm3^3 min−1^{-1}.

Section 4

Second order derivatives

Differentiating dydx\frac{\mathrm{d}y}{\mathrm{d}x} again gives the second derivative d2ydx2=f′′(x).\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=f''(x). It is the rate of change of the gradient. Where f′′(x)>0f''(x)>0 the gradient is increasing, and where f′′(x)<0f''(x)<0 it is decreasing. Example: f(x)=x3−6x2+9x+2f(x)=x^3-6x^2+9x+2 gives f′(x)=3x2−12x+9f'(x)=3x^2-12x+9 and f′′(x)=6x−12f''(x)=6x-12. Since f′′(x)>0f''(x)>0 for x>2x>2, the gradient increases for x>2x>2. The rule that xnx^n differentiates to nxn−1nx^{n-1} is covered in the next subtopic. The chain rule is not needed here.

Key termssecond derivative
Common mistake

Writing d2ydx2\frac{\mathrm{d}^2y}{\mathrm{d}x^2} as (dydx)2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2. They are different: the first differentiates twice, the second squares the gradient.

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Carry on to the next subtopic.

Exam questions on The derivative as a gradient and rate of change

  1. A curve has equation y=x2+3xy=x^2+3x. The point PP has coordinates (2,10)(2,10) and the point QQ on the curve has xx-coordinate 2+h2+h, where h≠0h\neq0.
    Explain why the gradient of the chord PQPQ is not equal to the gradient of the tangent at PP, and how the gradient of the tangent can be found from it.2 marks
  2. The volume VV cm3^3 of water in a tank at time tt minutes is modelled by V=40+6t−0.5t2V=40+6t-0.5t^2 for 0≤t≤100\le t\le10.
    Find the time at which the volume of water momentarily stops changing, and explain what the sign of dVdt\frac{\mathrm{d}V}{\mathrm{d}t} tells you about the volume just before this time.2 marks
  3. The curve y=f(x)y=f(x) has equation f(x)=x3−6x2+9x+2f(x)=x^3-6x^2+9x+2.
    Find f′(x)f'(x) and hence solve f′(x)=0f'(x)=0.3 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).