## Precalculus (6th Edition) Blitzer

The statement “If $f\left( x \right)=\pi {{x}^{2}}$ describes the area of a circle, $f\left( x \right),$ with radius x, $f'\left( 5 \right)>f'\left( 2 \right)$ because the area increases more rapidly as the radius increases” makes sense.
The derivative of the function $f\left( x \right)=\pi {{x}^{2}}$ at x is given by $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$, provided this limit exists. \begin{align} & f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\pi {{\left( x+h \right)}^{2}}-\pi {{x}^{2}}}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\pi \left( {{x}^{2}}+{{h}^{2}}+2xh \right)-\pi {{x}^{2}}}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\pi {{x}^{2}}+\pi {{h}^{2}}+2\pi xh-\pi {{x}^{2}}}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\pi {{h}^{2}}+2\pi xh}{h} \end{align} \begin{align} & =\underset{h\to 0}{\mathop{\lim }}\,\left( \pi h+2\pi x \right) \\ & =\pi \left( 0 \right)+2\pi x \\ & =2\pi x \\ \end{align} To find the value of $f'\left( 5 \right)$, substitute $x=5$ in $f'\left( x \right)=2\pi x$. $f'\left( 5 \right)=2\pi \left( 5 \right)=10\pi$ To find the value of $f'\left( 2 \right)$, substitute $x=2$ in $f'\left( x \right)=2\pi x$. $f'\left( 2 \right)=2\pi \left( 2 \right)=4\pi$ Thus $f'\left( 5 \right)>f'\left( 2 \right)$ Thus, the area increases more rapidly as the radius increases. Thus, the statement makes sense.