Calculus I - The Definition of the Derivative

Show Mobile Notice Show All Notes Hide All Notes

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 3.1 : The Definition of the Derivative

Back to Problem List

8. Use the definition of the derivative to find the derivative of,

\[R\left( z \right) = \frac{5}{z}\]

Show All Steps Hide All Steps

Start Solution

First we need to plug the function into the definition of the derivative.

\[R'\left( z \right) = \mathop {\lim }\limits_{h \to 0} \frac{{R\left( {z + h} \right) - R\left( z \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{5}{{z + h}} - \frac{5}{z}} \right)\]

Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems.

Also note that in order to make the problem a little easier to read rewrote the rational expression in the definition a little bit. This doesn’t need to be done, but will make things a little nicer to look at.

Show Step 2

Next we need to combine the two rational expressions into a single rational expression.

\[R'\left( z \right) = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{{5z - 5\left( {z + h} \right)}}{{z\left( {z + h} \right)}}} \right)\] Show Step 3

Now all that we need to do is some algebra and we’ll be done.

\[R'\left( z \right) = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{{5z - 5z - 5h}}{{z\left( {z + h} \right)}}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{{ - 5h}}{{z\left( {z + h} \right)}}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{ - 5}}{{z\left( {z + h} \right)}} = - \frac{5}{{{z^2}}}\]

The derivative for this function is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{R'\left( z \right) = - \frac{5}{{{z^2}}}}}\]