Chapter 4 - Calculating the Derivative - 4.2 Derivatives of Products and Quotients - 4.2 Exercises - Page 218: 54

The average number of vehicles waiting in a line to enter a parking ramp can be modeled by the function $$ f(x)=\frac{x^{2}}{2(1-x)}, $$ Then, by the quotient rule, $$ f^{\prime}(x)=\left(\frac{u(x)}{v(x)}\right)^{\prime}=\frac{v(x) \cdot u^{\prime}(x)-u(x) v^{\prime}(x)}{[v(x)]^{2}} $$ where $u(x)=x^{2}$ and $v(x)=2(1-x)$ , then the rate of change of the number of vehicles in line with respect to the traffic intensity $f^{\prime}(x) $ is given by $$ \begin{aligned} f^{\prime}(x) &=\frac{2(1-x)(2 x)-x^{2}(-2)}{[2(1-x)]^{2}} \\ &=\frac{4 x-4 x^{2}+2 x^{2}}{4(1-x)^{2}} \\ &=\frac{4 x-2 x^{2}}{4(1-x)^{2}} \\ &=\frac{2 x(2-x)}{4(1-x)^{2}} \\ &=\frac{x(2-x)}{2(1-x)^{2}} \end{aligned} $$ the rate of change of the number of vehicles in line with respect to the traffic intensity for the following values of the intensity. (a) $x=0.1$ $$ \begin{aligned} f^{\prime}(0.1) &=\frac{(0.1)(2-(0.1))}{2(1-(0.1))^{2}} \\ &\approx 0.1173 \end{aligned} $$ (b) $x=0.6$ $$ \begin{aligned} f^{\prime}(0.6) &=\frac{(0.6)(2-(0.6))}{2(1-(0.6))^{2}} \\ &\approx 2.625 \end{aligned} $$

The average number of vehicles waiting in a line to enter a parking ramp can be modeled by the function $$ f(x)=\frac{x^{2}}{2(1-x)}, $$ Then, by the quotient rule, $$ f^{\prime}(x)=\left(\frac{u(x)}{v(x)}\right)^{\prime}=\frac{v(x) \cdot u^{\prime}(x)-u(x) v^{\prime}(x)}{[v(x)]^{2}} $$ where $u(x)=x^{2}$ and $v(x)=2(1-x)$ , then the rate of change of the number of vehicles in line with respect to the traffic intensity $f^{\prime}(x) $ is given by $$ \begin{aligned} f^{\prime}(x) &=\frac{2(1-x)(2 x)-x^{2}(-2)}{[2(1-x)]^{2}} \\ &=\frac{4 x-4 x^{2}+2 x^{2}}{4(1-x)^{2}} \\ &=\frac{4 x-2 x^{2}}{4(1-x)^{2}} \\ &=\frac{2 x(2-x)}{4(1-x)^{2}} \\ &=\frac{x(2-x)}{2(1-x)^{2}} \end{aligned} $$ the rate of change of the number of vehicles in line with respect to the traffic intensity for the following values of the intensity. (a) $x=0.1$ $$ \begin{aligned} f^{\prime}(0.1) &=\frac{(0.1)(2-(0.1))}{2(1-(0.1))^{2}} \\ &\approx 0.1173 \end{aligned} $$ (b) $x=0.6$ $$ \begin{aligned} f^{\prime}(0.6) &=\frac{(0.6)(2-(0.6))}{2(1-(0.6))^{2}} \\ &\approx 2.625 \end{aligned} $$