Answer
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.
Work Step by Step
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.
