Find $f^\prime$ if $f( x) =x^{3}-1$. Then graph $y=f( x)$ and $y=f^\prime ( x)$ on the same set of coordinate axes.

Figure 9

Solution $f( x) =x^{3}-1$ so $$f( x+h) =( x+h) ^{3}-1=x^{3}+3hx^{2}+3h^{2}x+h^{3}-1$$ Using form (3), we find $$\begin{eqnarray*} f^\prime ( x) &=&\lim\limits_{h\rightarrow 0}\dfrac{f( x+h) -f( x) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{( x^{3}+3hx^{2}+3h^{2}x+h^{3}-1) -( x^{3}-1) }{h}\\[7pt] &=&\lim\limits_{h\rightarrow 0}\dfrac{3hx^{2}+3h^{2}x+h^{3}}{h} =\lim\limits_{h\rightarrow 0}\dfrac{h( 3x^{2}+3hx+h^{2}) }{h}\\[7pt] &=&\lim\limits_{h\rightarrow 0}( 3x^{2}{+}3hx{+}h^{2}) =3x^{2} \end{eqnarray*}$$

The graphs of $f$ and $f^\prime$ are shown in Figure 9.