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,

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