Second Derivatives (AQA GCSE Further Maths)

Revision Note

Last updated

5 October 2024

If you differentiate the derivative of a function (i.e. differentiate the function a second time) you get the second order derivative of the function
- The second order derivative can be referred to simply as the second derivative
We can write the second derivative as $\frac{d^{2} y}{d x^{2}}$
Note the position of the powers of 2
- differentiating twice (so $d^{2}$ ) with respect to $x$ twice (so $x^{2}$ )
A first derivative is the rate of change of a function (the gradient)
- a second order derivative is the rate of change of the rate of change of a function
  - i.e. the rate of change of the function’s gradient
- A positive second derivative means the gradient is increasing (graph is becoming steeper)
  - For instance in a u-shape, where the gradient is changing from negative to positive
- A negative second derivative means the gradient is decreasing (graph is becoming less steep)
  - For instance in an n-shape, where the gradient is changing from positive to negative
Second order derivatives can be used to test whether a point is a minimum or maximum
To find a second derivative, you simply differentiate twice!
- It is important to write down your working with the correct notation, so you know what each expression means
- For example
  - $y = 5 x^{3} + 10 x^{2}$
  - $\frac{d y}{d x} = 15 x^{2} + 20 x$
  - $\frac{d^{2} y}{d x^{2}} = 30 x + 20$

Even if you think you can find the second derivative in your head and write it down, make sure you write down the first derivative as well
- If you make a mistake, you will most likely get marks for finding the first derivative

Work out $\frac{d^{2} y}{d x^{2}}$ when

(a) $y = x^{5} - 2 x^{3} + 7 x^{2} + 9 x - 18$

Find the first derivative of the function first by considering each term in turn.

$\begin{array}{rcl} \frac{d y}{d x} & = & 5 x^{4} - 6 x^{2} + 14 x + 9 \end{array}$

Find the second derivative, $\frac{d^{2} y}{d x^{2}}$ by differentiating each term in the first derivative.

$\begin{array}{rcl} \frac{d^{2} y}{d x^{2}} & = & 20 x^{3} - 12 x + 14 \end{array}$

(b) $\begin{matrix} \end{matrix}$ $y = \frac{3 x + 7}{x^{4}}$

To find the first derivative of the function, begin by separating the terms in the fraction.

$\begin{array}{rcl} y & = & \frac{3 x}{x^{4}} + \frac{7}{x^{4}} \end{array}$

Rewrite each term using index notation so that they are in a form that can be differentiated.

$\begin{array}{rcl} y & = & 3 x (x^{- 4}) + 7 x^{- 4} = 3 x^{- 3} + 7 x^{- 4} \end{array}$

Find the first derivative of the function first by differentiating each term in turn.

$\begin{array}{rcl} \frac{d y}{d x} & = & - 9 x^{- 4} - 28 x^{- 5} \end{array}$

Find the second derivative, $\frac{d^{2} y}{d x^{2}}$ by differentiating each term in the first derivative.

$\begin{array}{rcl} \frac{d^{2} y}{d x^{2}} & = & - 9 (- 4) x^{- 5} - 28 (- 5) x^{- 6} = 36 x^{- 5} + 140 x^{- 6} \end{array}$

You can turn the second derivative back into the same format as the original function by rewriting as a fraction.

$\begin{array}{rcl} \frac{d^{2} y}{d x^{2}} & = & \frac{36}{x^{5}} + \frac{140}{x^{6}} = \frac{36 x + 140}{x^{6}} \end{array}$

$\begin{array}{rcl} \frac{d^{2} y}{d x^{2}} & = & \frac{36 x + 140}{x^{6}} \end{array}$

the (exam) results speak for themselves:

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