Chapter 4 - Calculating the Derivative - 4.3 The Chain Rule - 4.3 Exercises - Page 227: 62

The total number of bacteria (in millions) present in a culture is given by $$ \begin{aligned} N(t)&=2t(5t+9)^{\frac{1}{2}}+12 \end{aligned} $$ Then, the rate of change of the population of bacteria with respect to time is given by $$ \begin{aligned} N^{\prime}(t) &=(2 t)\left[\frac{1}{2}(5 t+9)^{-1 / 2}(5)\right] +2(5 t+9)^{1 / 2}+0 \\ &= 5 t(5 t+9)^{-1 / 2}+2(5 t+9)^{1 / 2} \\ &=(5 t+9)^{-1 / 2}[5 t+2(5 t+9)] \\ &=(5 t+9)^{-1 / 2}(15 t+18) \\ &= \frac{15 t+18}{(5 t+9)^{1 / 2}} \end{aligned} $$ Now, the rate of change of the population of bacteria with respect to time for the following numbers of hours. (a) 0 hours $$ \begin{aligned} N^{\prime}(0) &= \frac{15 (0)+18}{(5 (0)+9)^{1 / 2}} \\ &=\frac{18}{(9)^{1 / 2}}\\ &=\frac{18}{3}\\ &=6 \end{aligned} $$ (b) $\frac{7}{5}$ hours $$ \begin{aligned} N^{\prime}(\frac{7}{5}) &= \frac{15 (\frac{7}{5})+18}{(5 (\frac{7}{5})+9)^{1 / 2}} \\ &=\frac{21+18}{(7+9)^{1 / 2}}\\ &=\frac{39}{(16)^{1 / 2}}\\ &=\frac{39}{4}\\ &=9.75 \end{aligned} $$ (c) 8 hours $$ \begin{aligned} N^{\prime}(8) &= \frac{15 (8)+18}{(5 (8)+9)^{1 / 2}} \\ &=\frac{120+18}{(49)^{1 / 2}}\\ &=\frac{138}{7}\\ & \approx 19.71 \end{aligned} $$

The total number of bacteria (in millions) present in a culture is given by $$ \begin{aligned} N(t)&=2t(5t+9)^{\frac{1}{2}}+12 \end{aligned} $$ Then, the rate of change of the population of bacteria with respect to time is given by $$ \begin{aligned} N^{\prime}(t) &=(2 t)\left[\frac{1}{2}(5 t+9)^{-1 / 2}(5)\right] +2(5 t+9)^{1 / 2}+0 \\ &= 5 t(5 t+9)^{-1 / 2}+2(5 t+9)^{1 / 2} \\ &=(5 t+9)^{-1 / 2}[5 t+2(5 t+9)] \\ &=(5 t+9)^{-1 / 2}(15 t+18) \\ &= \frac{15 t+18}{(5 t+9)^{1 / 2}} \end{aligned} $$ Now, the rate of change of the population of bacteria with respect to time for the following numbers of hours. (a) 0 hours $$ \begin{aligned} N^{\prime}(0) &= \frac{15 (0)+18}{(5 (0)+9)^{1 / 2}} \\ &=\frac{18}{(9)^{1 / 2}}\\ &=\frac{18}{3}\\ &=6 \end{aligned} $$ (b) $\frac{7}{5}$ hours $$ \begin{aligned} N^{\prime}(\frac{7}{5}) &= \frac{15 (\frac{7}{5})+18}{(5 (\frac{7}{5})+9)^{1 / 2}} \\ &=\frac{21+18}{(7+9)^{1 / 2}}\\ &=\frac{39}{(16)^{1 / 2}}\\ &=\frac{39}{4}\\ &=9.75 \end{aligned} $$ (c) 8 hours $$ \begin{aligned} N^{\prime}(8) &= \frac{15 (8)+18}{(5 (8)+9)^{1 / 2}} \\ &=\frac{120+18}{(49)^{1 / 2}}\\ &=\frac{138}{7}\\ & \approx 19.71 \end{aligned} $$