Calculus I - The Definition of the Derivative

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Section 3.1 : The Definition of the Derivative

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

\[V\left( t \right) = \frac{{t + 1}}{{t + 4}}\]

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

\[V'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{{V\left( {t + h} \right) - V\left( t \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{{t + h + 1}}{{t + h + 4}} - \frac{{t + 1}}{{t + 4}}} \right)\]

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.

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

\[V'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{{\left( {t + h + 1} \right)\left( {t + 4} \right) - \left( {t + 1} \right)\left( {t + h + 4} \right)}}{{\left( {t + h + 4} \right)\left( {t + 4} \right)}}} \right)\] Show Step 3

Now all that we need to do is some algebra (and it will get a little messy here, but that is somewhat common with these types of problems) and we’ll be done.

\[\begin{align*}V'\left( t \right) & = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{{{t^2} + th + 5t + 4h + 4 - \left( {{t^2} + th + 5t + h + 4} \right)}}{{\left( {t + h + 4} \right)\left( {t + 4} \right)}}} \right)\\ & = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{{{t^2} + th + 5t + 4h + 4 - {t^2} - th - 5t - h - 4}}{{\left( {t + h + 4} \right)\left( {t + 4} \right)}}} \right)\\ & = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{{3h}}{{\left( {t + h + 4} \right)\left( {t + 4} \right)}}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{3}{{\left( {t + h + 4} \right)\left( {t + 4} \right)}} = \frac{3}{{{{\left( {t + 4} \right)}^2}}}\end{align*}\]

The derivative for this function is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{V'\left( t \right) = \frac{3}{{{{\left( {t + 4} \right)}^2}}}}}\]