Chapter 4 - Calculating the Derivative - 4.3 The Chain Rule - 4.3 Exercises - Page 227: 63

the amount of the calcium remaining in the bloodstream (in milligrams per cubic centimeter) $t$ days after the initial injection is approximated by $$ \begin{aligned} C(t)&=\frac{1}{2}(2t+1)^{-\frac{1}{2}} \end{aligned} $$ Then, the rate of change of the population of bacteria with respect to time is given by $$ \begin{aligned} C^{\prime}(t) &=\frac{1}{2}(-\frac{1}{2})(2t+1)^{-\frac{-3}{2}} (2)\\ &=(-\frac{1}{2})(2t+1)^{-\frac{-3}{2}} \\ \end{aligned} $$ Now, the rate of change of the calcium level with respect to time for the following numbers of days. (a) 0 day $$ \begin{aligned} C^{\prime}(0) &= (-\frac{1}{2})(2(0)+1)^{-\frac{-3}{2}}\\ &=-\frac{1}{2}\\ \end{aligned} $$ (b) $4$ days $$ \begin{aligned} C^{\prime}(4) &= (-\frac{1}{2})(2(4)+1)^{-\frac{-3}{2}}\\ &= (-\frac{1}{2})(9)^{-\frac{-3}{2}}\\ &= -\frac{1}{2(\sqrt {9})^{3}}\\ &=-\frac{1}{54}\\ & \approx -0.02 \end{aligned} $$ (c) $7.5$ days $$ \begin{aligned} C^{\prime}(7.5) &= (-\frac{1}{2})(2(7.5)+1)^{-\frac{-3}{2}}\\ &= (-\frac{1}{2})(16)^{-\frac{-3}{2}}\\ &= -\frac{1}{2(\sqrt {16})^{3}}\\ &= -\frac{1}{2(64)}\\ &=-\frac{1}{128}\\ & \approx -0.008 \end{aligned} $$ (d) Since $$ C^{\prime}(t)=(-\frac{1}{2})(2t+1)^{-\frac{-3}{2}} \lt 0 $$ for all $t \geq 0 $, so C always decreasing.

the amount of the calcium remaining in the bloodstream (in milligrams per cubic centimeter) $t$ days after the initial injection is approximated by $$ \begin{aligned} C(t)&=\frac{1}{2}(2t+1)^{-\frac{1}{2}} \end{aligned} $$ Then, the rate of change of the population of bacteria with respect to time is given by $$ \begin{aligned} C^{\prime}(t) &=\frac{1}{2}(-\frac{1}{2})(2t+1)^{-\frac{-3}{2}} (2)\\ &=(-\frac{1}{2})(2t+1)^{-\frac{-3}{2}} \\ \end{aligned} $$ Now, the rate of change of the calcium level with respect to time for the following numbers of days. (a) 0 day $$ \begin{aligned} C^{\prime}(0) &= (-\frac{1}{2})(2(0)+1)^{-\frac{-3}{2}}\\ &=-\frac{1}{2}\\ \end{aligned} $$ (b) $4$ days $$ \begin{aligned} C^{\prime}(4) &= (-\frac{1}{2})(2(4)+1)^{-\frac{-3}{2}}\\ &= (-\frac{1}{2})(9)^{-\frac{-3}{2}}\\ &= -\frac{1}{2(\sqrt {9})^{3}}\\ &=-\frac{1}{54}\\ & \approx -0.02 \end{aligned} $$ (c) $7.5$ days $$ \begin{aligned} C^{\prime}(7.5) &= (-\frac{1}{2})(2(7.5)+1)^{-\frac{-3}{2}}\\ &= (-\frac{1}{2})(16)^{-\frac{-3}{2}}\\ &= -\frac{1}{2(\sqrt {16})^{3}}\\ &= -\frac{1}{2(64)}\\ &=-\frac{1}{128}\\ & \approx -0.008 \end{aligned} $$ (d) Since $$ C^{\prime}(t)=(-\frac{1}{2})(2t+1)^{-\frac{-3}{2}} \lt 0 $$ for all $t \geq 0 $, so C always decreasing.