Differentiation Quiz

Each question shows a cubic function. Choose the correct derivative.

Question
Score: 0 / 0

Calculus — Ordinary Level (Examples & Practice)

Max/min questions use quadratics only. For cubics, you may be asked for slope or tangent elsewhere (not on this sheet).

1

Differentiate by rule

Use the power rule: \(\dfrac{\mathrm{d}}{\mathrm{d}x}(x^n)=nx^{\,n-1}\). Differentiate term-wise; constants vanish.

Example

Find \(f'(x), g'(x), h'(x)\), and the second derivative \(H''(x)\):

\[ \text{(i) } f(x)=3x+7,\quad \text{(ii) } g(x)=2x^2-5x+1,\quad \text{(iii) } h(x)=x^3-6x^2+4x-9,\quad H(x)=4x^3-3x^2+2x-5. \]

  • \(f'(x)=3\)
  • \(g'(x)=4x-5\)
  • \(h'(x)=3x^2-12x+4\)
  • \(H'(x)=12x^2-6x+2\Rightarrow H''(x)=24x-6\)

Practice

\[ p(x)=5x-4,\quad q(x)=x^2+6x-3,\quad r(x)=2x^3+x^2-7x+8. \] Find \(p'(x),\,q'(x),\,r'(x)\) and \(r''(x)\).

\(p'(x)=5,\quad q'(x)=2x+6,\quad r'(x)=6x^2+2x-7,\quad r''(x)=12x+2.\)

2

Slopes and tangent lines

Slope at \(x=a\) is \(\left.\dfrac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=a}\). Tangent: \(y-y_1=m(x-x_1)\).

Example

Let \(y=x^2+2x-3\). Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) at \(x=2\) and the tangent equation there.

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x+2\Rightarrow m=6\) at \(x=2\). Point: \(y(2)=5\Rightarrow(2,5)\). Tangent: \(y-5=6(x-2)\Rightarrow y=6x-7\).

Practice

For the function \(y=x^3-x\), find the slope at \(x=1\) and the tangent equation there.

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=3x^2-1\Rightarrow m=2\) at \(x=1\). Point \((1,0)\). Tangent: \(y=2x-2\).

3

Rates of change

Interpret derivatives as rates: \(v(t)=\dfrac{\mathrm{d}s}{\mathrm{d}t}\), \(a(t)=\dfrac{\mathrm{d}v}{\mathrm{d}t}\); or \(\dfrac{\mathrm{d}}{\mathrm{d}x}\) of a formula.

Example

The distance is \(s=4t^2+2t\) m after \(t\) s. Find \(v(t)\), \(a(t)\), and \(v(3)\).

\(v(t)=\dfrac{\mathrm{d}s}{\mathrm{d}t}=8t+2,\quad a(t)=\dfrac{\mathrm{d}v}{\mathrm{d}t}=8.\) Hence \(v(3)=26\,\text{m s}^{-1}\).

Practice

Area of a circle: \(A=\pi r^2\). Find \(\dfrac{\mathrm{d}A}{\mathrm{d}r}\) and evaluate at \(r=5\).

\(\dfrac{\mathrm{d}A}{\mathrm{d}r}=2\pi r\Rightarrow 10\pi\) at \(r=5\).

4

Minimum value & minimum point

Solve \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\) to get the turning point \(x\)-value, then substitute to find the minimum value and the minimum point. (No use of the second derivative is required.)

Example

Find the minimum value of \(y=x^2-6x+5\) and the corresponding minimum point.

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x-6=0\Rightarrow x=3\). Minimum value \(=y(3)=9-18+5=-4\). Minimum point \(=(3,-4)\).

Practice

Let \(h(t)=t^2-4t+7\). Find the minimum value of \(h(t)\) and the minimum point \((t,h(t))\).

\(\dfrac{\mathrm{d}h}{\mathrm{d}t}=2t-4=0\Rightarrow t=2\). Minimum value \(=h(2)=4-8+7=3\). Minimum point \(=(2,3)\).

Video Tutorials

Slope and Equation of a Tangent

Find the slope of the tangent to the curve \[ y = 3x^{2} + 4x - 5 \] at the point \((1,2)\). Hence, find the equation of this tangent.
Show video
Show worked solution
1. Differentiate to find the slope function.
For \(y = 3x^{2} + 4x - 5\): \[ \frac{dy}{dx} = 6x + 4. \]
2. Find the slope at \(x = 1\).
\[ \left.\frac{dy}{dx}\right|_{x=1} = 6(1) + 4 = 10. \] So, the slope of the tangent at \((1,2)\) is \(m = 10\).
3. Use the point–slope form of a line.
A line with slope \(m\) through \((x_{1},y_{1})\) is \[ y - y_{1} = m(x - x_{1}). \] Here, \(m = 10\) and \((x_{1},y_{1}) = (1,2)\): \[ y - 2 = 10(x - 1). \]
4. Simplify to slope–intercept form.
\[ y - 2 = 10x - 10 \quad\Rightarrow\quad y = 10x - 8. \] So the equation of the tangent is \[ y = 10x - 8. \]

Displacement, Velocity and Acceleration

A body moves in a straight line so that its distance \(s\) metres from a fixed point after \(t\) seconds is given by \[ s = 10 + 27t - t^{3}. \]

Find:

  1. the speed after 2 seconds,
  2. after how many seconds the body is at rest,
  3. the acceleration after 2 seconds.
Show video
Show worked solution
1. Differentiate \(s\) to get velocity.
\[ v = \frac{ds}{dt} = 27 - 3t^{2}. \]
(i) Speed after 2 seconds
\[ v = 27 - 3(2)^{2} = 27 - 12 = 15. \] Therefore, the speed is \(15~\text{m s}^{-1}\).
(ii) Body at rest
Body is at rest when \(v = 0\): \[ 27 - 3t^{2} = 0 \Rightarrow t^{2} = 9 \Rightarrow t = 3. \] Hence, the body is at rest after \(3\) seconds.
2. Differentiate \(v\) to get acceleration.
\[ a = \frac{dv}{dt} = -6t. \]
(iii) Acceleration after 2 seconds
\[ a = -6(2) = -12. \] Therefore, the acceleration is \(-12~\text{m s}^{-2}\).

Finding the Minimum Value

A factory produces \(n\) items per hour. The overhead costs are given by \[ C(n) = €\left(400 - 16n + \frac{n^{2}}{4}\right). \]

How many items should be produced to keep the overhead costs to a minimum?

Show video
Show worked solution
1. Differentiate \(C(n)\) with respect to \(n\).
\[ \frac{dC}{dn} = -16 + \frac{n}{2}. \]
2. Set derivative to zero for stationary points.
\[ -16 + \frac{n}{2} = 0 \Rightarrow \frac{n}{2} = 16 \Rightarrow n = 32. \]
3. Check the second derivative.
\[ \frac{d^{2}C}{dn^{2}} = \frac{1}{2} > 0. \] Therefore, the function has a minimum at \(n = 32\).
4. Minimum overhead cost.
Substituting \(n = 32\): \[ C(32) = €(400 - 16(32) + \frac{32^{2}}{4}) = €(400 - 512 + 256) = €144. \]
Answer: The overhead cost is minimised when 32 items are produced per hour, giving a minimum cost of €144.