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### Section 3-1 : The Definition of the Derivative

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

$Z\left( t \right) = \sqrt {3t - 4}$

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

$Z'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{{Z\left( {t + h} \right) - Z\left( t \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {3\left( {t + h} \right) - 4} - \sqrt {3t - 4} }}{h}$

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

Show Step 2

Next we need to rationalize the numerator.

$Z'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\sqrt {3\left( {t + h} \right) - 4} - \sqrt {3t - 4} } \right)}}{h}\frac{{\left( {\sqrt {3\left( {t + h} \right) - 4} + \sqrt {3t - 4} } \right)}}{{\left( {\sqrt {3\left( {t + h} \right) - 4} + \sqrt {3t - 4} } \right)}}$ 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*}Z'\left( t \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{3\left( {t + h} \right) - 4 - \left( {3t - 4} \right)}}{{h\left( {\sqrt {3\left( {t + h} \right) - 4} + \sqrt {3t - 4} } \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{3t + 3h - 4 - 3t + 4}}{{h\left( {\sqrt {3\left( {t + h} \right) - 4} + \sqrt {3t - 4} } \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{3h}}{{h\left( {\sqrt {3\left( {t + h} \right) - 4} + \sqrt {3t - 4} } \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{3}{{\sqrt {3\left( {t + h} \right) - 4} + \sqrt {3t - 4} }} = \frac{3}{{2\sqrt {3t - 4} }}\end{align*}

Be careful when multiplying out the numerator here. It is easy to lose track of the minus sign (or parenthesis for that matter) on the second term. This is a very common mistake that students make.

The derivative for this function is then,

$\require{bbox} \bbox[2pt,border:1px solid black]{{Z'\left( t \right) = \frac{3}{{2\sqrt {3t - 4} }}}}$