Next we need to rationalize the numerator.

\[f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\sqrt {1 - 9\left( {x + h} \right)} - \sqrt {1 - 9x} } \right)}}{h}\frac{{\left( {\sqrt {1 - 9\left( {x + h} \right)} + \sqrt {1 - 9x} } \right)}}{{\left( {\sqrt {1 - 9\left( {x + h} \right)} + \sqrt {1 - 9x} } \right)}}\] 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.

\[\begin{align*}f'\left( x \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{1 - 9\left( {x + h} \right) - \left( {1 - 9x} \right)}}{{h\left( {\sqrt {1 - 9\left( {x + h} \right)} + \sqrt {1 - 9x} } \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{1 - 9x - 9h - 1 + 9x}}{{h\left( {\sqrt {1 - 9\left( {x + h} \right)} + \sqrt {1 - 9x} } \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{ - 9h}}{{h\left( {\sqrt {1 - 9\left( {x + h} \right)} + \sqrt {1 - 9x} } \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - 9}}{{\sqrt {1 - 9\left( {x + h} \right)} + \sqrt {1 - 9x} }} = \frac{{ - 9}}{{2\sqrt {1 - 9x} }}\end{align*}\]

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,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{f'\left( x \right) = \frac{{ - 9}}{{2\sqrt {1 - 9x} }}}}\]