## Precalculus (6th Edition) Blitzer

The instantaneous rate of change of the surface area with respect to the radius is $48\pi \text{ square inches}$ per inch when $x=6\text{ inches}$.
Consider that the surface area of a sphere is described by the function $f\left( x \right)=4\pi {{x}^{2}}$ Compute the derivative of $f\left( x \right)=4\pi {{x}^{2}}$ using the formula $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$ as follows: To compute $f\left( x+h \right)$, substitute $x=x+h$ in the function $f\left( x \right)={{x}^{2}}$. \begin{align} & f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{4\pi {{\left( x+h \right)}^{2}}-4\pi {{x}^{2}}}{h} \end{align} Now, simplify $4\pi {{\left( x+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$ \begin{align} & f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{4\pi \left( {{x}^{2}}+2xh+{{h}^{2}} \right)-4\pi {{x}^{2}}}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{4\pi {{x}^{2}}+8\pi xh+4\pi {{h}^{2}}-4\pi {{x}^{2}}}{h} \end{align} Combine the like terms in the numerator; then divide the numerator and denominator by h; this gives, $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 8\pi x+4\pi h \right)$ Apply the limits, \begin{align} & f'\left( x \right)=8\pi x+4\pi \cdot 0 \\ & =8\pi x \end{align} Now, substitute $x=6$ in $f'\left( x \right)$ to compute the instantaneous change of “f” at $6$. \begin{align} & f'\left( 6 \right)=8\pi \cdot 6 \\ & =48\pi \end{align} Thus, the instantaneous rate of change of the surface area with respect to the radius is $48\pi \text{ square inches}$ per inch when $x=6\text{ inches}$.