2 First derivatives

Our aim is to approximate the slope of a curve $f$ at a particular point $x=a$ in terms of $f\left(a\right)$ and the value of $f$ at a nearby point where $x=a+h$ . The shorter broken line Figure 11 may be thought of as giving a reasonable approximation to the required slope (shown by the longer broken line), if $h$ is small enough.

So we might approximate

$f^{\prime }\left(a\right)\approx \text{slope of short broken line}=\frac{\text{difference in the }y\text{-values }}{\text{difference in the }x\text{-values}}=\frac{f\left(a+h\right)-f\left(a\right)}{h}.$

This is called a one-sided difference or forward difference approximation to the derivative of $f$ .

A second version of this arises on considering a point to the left of $a$ , rather than to the right as we did above. In this case we obtain the approximation

$f^{\prime }\left(a\right)\approx \frac{f\left(a\right)-f\left(a-h\right)}{h}$

This is another one-sided difference , called a backward difference , approximation to $f^{\prime }\left(a\right)$ .

A third method for approximating the first derivative of $f$ can be seen in Figure 12.

Here we approximate as follows

$f^{\prime }\left(a\right)\approx \text{slope of short broken line}=\frac{\text{difference in the }y\text{-values }}{\text{difference in the }x\text{-values}}=\frac{f\left(x+h\right)-f\left(x-h\right)}{2h}$

This is called a central difference approximation to $f^{\prime }\left(a\right)$ .

These first, rather artificial, examples will help fix our ideas before we move on to more realistic applications.

Example 18

Use a forward difference, and the values of $h$ shown, to approximate the derivative of $cos\left(x\right)$ at $x=\pi ∕3$ .

1. $h=0.1$
2. $h=0.01$
3. $h=0.001$
4. $h=0.0001$

Work to 8 decimal places throughout.

Solution

1. $f^{\prime }\left(a\right)\approx \frac{cos\left(a+h\right)-cos\left(a\right)}{h}=\frac{0.41104381-0.5}{0.1}=-0.88956192$
   
   
2. $f^{\prime }\left(a\right)\approx \frac{cos\left(a+h\right)-cos\left(a\right)}{h}=\frac{0.49131489-0.5}{0.01}=-0.86851095$
   
   
3. $f^{\prime }\left(a\right)\approx \frac{cos\left(a+h\right)-cos\left(a\right)}{h}=\frac{0.49913372-0.5}{0.001}=-0.86627526$
   
   
4. $f^{\prime }\left(a\right)\approx \frac{cos\left(a+h\right)-cos\left(a\right)}{h}=\frac{0.49991339-0.5}{0.0001}=-0.86605040$

One advantage of doing a simple example first is that we can compare these approximations with the ‘exact’ value which is

$f^{\prime }\left(a\right)=-sin\left(\pi ∕3\right)=-\frac{\sqrt{3}}{2}=-0.86602540\text{to 8 d.p.}$

Note that the accuracy levels of the four approximations in Example 15 are:

1. 1 d.p.
2. 2 d.p.
3. 3 d.p.
4. 3 d.p. (almost 4 d.p.)

   The errors to 6 d.p. are:

1. 0.023537
2. 0.002486
3. 0.000250
4. 0.000025

   Notice that the errors reduce by about a factor of 10 each time.

Example 19

Use a central difference, and the value of $h$ shown, to approximate the derivative of $cos\left(x\right)$ at $x=\pi ∕3$ .

1. $h=0.1$
2. $h=0.01$
3. $h=0.001$
4. $h=0.0001$

Work to 8 decimal places throughout.

Solution

1. $f^{\prime }\left(a\right)\approx \frac{cos\left(a+h\right)-cos\left(a-h\right)}{2h}=\frac{0.41104381-0.58396036}{0.2}=-0.86458275$
   
   
2. $f^{\prime }\left(a\right)\approx \frac{cos\left(a+h\right)-cos\left(a-h\right)}{2h}=\frac{0.49131489-0.50863511}{0.02}=-0.86601097$
   
   
3. $f^{\prime }\left(a\right)\approx \frac{cos\left(a+h\right)-cos\left(a-h\right)}{2h}=\frac{0.49913372-0.50086578}{0.002}=-0.86602526$
   
   
4. $f^{\prime }\left(a\right)\approx \frac{cos\left(a+h\right)-cos\left(a-h\right)}{2h}=\frac{0.49991339-0.50008660}{0.0002}=-0.86602540$

This time successive approximations generally have two extra accurate decimal places indicating a superior formula. This is illustrated again in the following Task.

Task!

Let $f\left(x\right)=ln\left(x\right)$ and $a=3$ . Using both a forward difference and a central difference, and working to 8 decimal places, approximate $f^{\prime }\left(a\right)$ using $h=0.1$ and $h=0.01$ .

(Note that this is another example where we can work out the exact answer, which in this case is $\frac{1}{3}$ .)

Answer

There is clearly little point in studying this technique if all we ever do is approximate quantities we could find exactly in another way. The following example is one in which this so-called differencing method is the best approach.

Example 20

The distance $x$ of a runner from a fixed point is measured (in metres) at intervals of half a second. The data obtained are

$\begin{array}{cccccc}\hfill t\hfill & \hfill 0.0\hfill & \hfill 0.5\hfill & \hfill 1.0\hfill & \hfill 1.5\hfill & \hfill 2.0\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill x\hfill & \hfill 0.00\hfill & \hfill 3.65\hfill & \hfill 6.80\hfill & \hfill 9.90\hfill & \hfill 12.15\hfill \end{array}$

Use central differences to approximate the runner’s velocity at times $t=0.5$ s and $t=1.25$ s.

Solution

Our aim here is to approximate $x^{\prime }\left(t\right)$ . The choice of $h$ is dictated by the available data given in the table.

Using data with $t=0.5$ s at its centre we obtain

$x^{\prime }\left(0.5\right)\approx \frac{x\left(1.0\right)-x\left(0.0\right)}{2\times 0.5}=6.80{\text{m s}}^{-1}.$

Data centred at $t=1.25$ s gives us the approximation

$x^{\prime }\left(1.25\right)\approx \frac{x\left(1.5\right)-x\left(1.0\right)}{2\times 0.25}=6.20{\text{m s}}^{-1}.$

Note the value of $h$ used.

Task!

The velocity $v$ (in $\text{m}{\text{s}}^{-1}$ ) of a rocket measured at half second intervals is

$\begin{array}{cccccc}\hfill t\hfill & \hfill 0.0\hfill & \hfill 0.5\hfill & \hfill 1.0\hfill & \hfill 1.5\hfill & \hfill 2.0\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill v\hfill & \hfill 0.000\hfill & \hfill 11.860\hfill & \hfill 26.335\hfill & \hfill 41.075\hfill & \hfill 59.051\hfill \end{array}$

Use central differences to approximate the acceleration of the rocket at times $t=1.0$ s and $t=1.75$ s.

Answer