Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 11

a.) Recall that the area of the square is $A = x^2$ where $x$ is the lenght of the sides of the square.

$\displaystyle A'(x) = \frac{d}{dx} (x^2) $
$\displaystyle A'(x) = 2x$

when $x = 15mm$
$\displaystyle A'(15) = 2 (15)$
$\displaystyle A'(15) = 30 \frac{mm^2}{mm}$


$A'(15)$ represents the rate at which the area is increasing with respect to the side of length x as it approach $15mm$.

b.) Recall that the perimeter is the sum of all sides, for square, we have $P(x) = 4x$, so $\displaystyle x = \frac{P(x)}{4}$ and we know that the rate of change of the area of the square is $A'(x) = 2x$
Hence,

$
\begin{equation}
\begin{aligned}
A'(x) &= 2 \left( \frac{P(x)}{4}\right) = \frac{P(x)}{2}\\
\\
A'(x) &= \frac{P(x)}{2}
\end{aligned}
\end{equation}
$





Hence, the resulting change in area is
$\Delta A = 2x (\Delta x) + (\Delta x)^2$
if $\Delta x$ is negligible,
$\Delta A \approx 2x \Delta x$