I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 3.1 : The Definition of the Derivative
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 SolutionFirst 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 2Next 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 3Now 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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{R'\left( z \right) = - \frac{5}{{{z^2}}}}}\]