Section 3.1 : The Definition of the Derivative
6. Use the definition of the derivative to find the derivative of,
\[f\left( x \right) = 2{x^3} - 1\]Show All Steps Hide All Steps
Start SolutionFirst we need to plug the function into the definition of the derivative.
\[f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{2{{\left( {x + h} \right)}^3} - 1 - \left( {2{x^3} - 1} \right)}}{h}\]Make sure that you properly evaluate the first function evaluation. This is one of the more common errors that students make with these problems.
Also watch for the parenthesis on the second function evaluation. You are subtracting off the whole function and so you need to make sure that you deal with the minus sign properly. Either put in the parenthesis as we’ve done here or make sure the minus sign get distributed through properly. This is another very common error and one that if you make will often make the problem impossible to complete.
Show Step 2Now all that we need to do is some algebra (and it might get a little messy here, but that is somewhat common with these types of problems) and we’ll be done.
\[\begin{align*}f'\left( x \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{2\left( {{x^3} + 3{x^2}h + 3x{h^2} + {h^3}} \right) - 1 - 2{x^3} + 1}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{2{x^3} + 6{x^2}h + 6x{h^2} + 2{h^3} - 1 - 2{x^3} + 1}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {6{x^2} + 6xh + 2{h^2}} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \left( {6{x^2} + 6xh + 2{h^2}} \right) = 6{x^2}\end{align*}\]The derivative for this function is then,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{f'\left( x \right) = 6{x^2}}}\]