Section 3.1 : The Definition of the Derivative
11. Use the definition of the derivative to find the derivative of,
\[f\left( x \right) = \sqrt {1 - 9x} \]Show All Steps Hide All Steps
Start SolutionFirst we need to plug the function into the definition of the derivative.
\[f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {1 - 9\left( {x + h} \right)} - \sqrt {1 - 9x} }}{h}\]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 2Next 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 3Now 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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{f'\left( x \right) = \frac{{ - 9}}{{2\sqrt {1 - 9x} }}}}\]