Calculus Exercise (3) : Rate of Change

General Rules of Differentiation
1	$\dfrac{d C}{d x}=0, C=$ constant
2	$\dfrac{d}{d x} x^{n}=n x^{n-1}$
3	$\dfrac{d}{d x}[u(x) \pm v(x)]=\dfrac{d}{d x}[u(x)] \pm \dfrac{d}{d x}[v(x)]$
4	$\dfrac{d}{d x}[C \cdot u(x)]=C \dfrac{d}{d x} u(x), C=$ constant

1. Differentiate the following with respect to $x$.
   
   (a) $4 x^{3}$
   
   (b) $\dfrac{4}{x^{3}}$
   
   (c) $\dfrac{2}{\sqrt[3]{x}}$
   
   (d) $x^{3}+\dfrac{1}{\sqrt{x}}$
   
   (e) $x^{2}-\dfrac{1}{x}-\dfrac{3}{x^{2}}$
   
   (f) $\dfrac{3 x^{2}-4 \sqrt{x}+1}{x}$
   
   (g) $(x+1)(x+2)$
   
   (h) $(3 x+1)(2-x)$
   
   (i) $(3 x-2)^{2}$
   
   (j) $(2 x+1)^{3}$
   
   (k) $\dfrac{2 x^{3}-3 x^{2}}{4 \sqrt{x}}$
   
   (l) $x^{3}-2 x+\dfrac{3}{\sqrt{x}}$
   
   





$\begin{aligned} \text { (a) }\quad &\frac{d}{d x}\left(4 x^{3}\right)\\\\ =&12 x\\\\ \text { (b) }\quad &\frac{d}{d x}\left(\frac{4}{x^{3}}\right)\\\\ =&\frac{d}{d x}\left(4 x^{-3}\right)\\\\ =&-12 x^{-4}\\\\ =&-\frac{12}{x^{4}}\\\\ \text { (c) }\quad &\frac{d}{d x}\left(\frac{2}{\sqrt[3]{x}}\right)\\\\ =&\frac{d}{d x}\left(2 x^{-\frac{1}{3}}\right)\\\\ =&-\frac{2}{3} x^{-\frac{4}{3}}\\\\ =&-\frac{2}{3 x^{\frac{4}{3}}}\\\\ \text { (d) }\quad &\frac{d}{d x}\left(x^{3}+\frac{1}{\sqrt{x}}\right)\\\\ =&\frac{d}{d x}\left(x^{3}\right)+\frac{d}{d x}\left(x^{-\frac{1}{2}}\right)\\\\ =&3 x^{2}-\frac{1}{2} x^{-\frac{3}{2}}\\\\ =&3 x^{2}-\frac{1}{2 x^{\frac{3}{2}}}\\\\ \text { (e) }\quad &\frac{d}{d x}\left(x^{2}-\frac{1}{x}-\frac{3}{x^{2}}\right) \\\\ =&\frac{d}{d x}\left(x^{2}\right)-\frac{d}{d x}\left(x^{-1}\right)-\frac{d}{d x}\left(3 x^{-2}\right) \\\\ =&2 x+x^{-2}+6 x^{-3} \\\\ =&2 x+\frac{1}{x^{2}}+\frac{6}{x^{3}} \\\\ \text { (f) }\quad &\frac{d}{d x}\left(\frac{3 x^{2}-4 \sqrt{x}+1}{x}\right)\\\\ =&\frac{d}{d x}\left(\frac{3 x^{2}}{x}-\frac{4 \sqrt{x}}{x}+\frac{1}{x}\right) \\\\ =&\frac{d}{d x}\left(3 x-4 x^{-\frac{1}{2}}+x^{-1}\right) \\\\ =&3+2 x^{-\frac{3}{2}}-x^{-2} \\\\ =&3+\frac{2}{x^{\frac{3}{2}}}-\frac{1}{x^{2}}\\\\ \text { (g) }\quad &\frac{d}{d x}[(x+1)(x+2)]\\\\ =&\frac{d}{d x}\left(x^{2}+3 x+2\right)\\\\ =&2 x+3\\\\ \text { (h) }\quad &\frac{d}{d x}[(3 x+1)(2-x)]\\\\ =&\frac{d}{d x}\left(2+5 x-3 x^{2}\right)\\\\ =&5-6 x\\\\ \text { (i) }\quad &\frac{d}{d x}(3 x-2)^{2}\\\\ =&\frac{d}{d x}\left(9 x^{2}-12 x+4\right)\\\\ =&18 x-12\\\\ \text { (j) }\quad &\frac{d}{d x}(2 x+1)^{3}\\\\ =&\frac{d}{d x}\left[(2 x)^{3}+3(2 x)^{2}+3(2 x)+1\right]\\\\ =&\frac{d}{d x}\left[8 x^{3}+12 x^{2}+6 x+1\right] \\\\ =&24 x^{2}+24 x+6\\\\ \text { (k) }\quad &\frac{d}{d x}\left[\frac{2 x^{3}-3 x^{2}}{4 \sqrt{x}}\right]\\\\ =&\frac{d}{d x}\left[\frac{2 x^{3}}{4 \sqrt{x}}-\frac{3 x^{2}}{4 \sqrt{x}}\right]\\\\ =&\frac{d}{d x}\left[\frac{1}{2} x^{\frac{5}{2}}-\frac{3}{4} x^{\frac{3}{2}}\right]\\\\ =&\frac{5}{4} x^{\frac{3}{2}}-\frac{9}{8} x^{\frac{1}{2}}\\\\ \text { (l) }\quad & \frac{d}{d x}\left[x^{3}-2 x+\frac{3}{\sqrt{x}}\right]\\\\ =&\frac{d}{d x}\left(x^{3}\right)-\frac{d}{d x}(2 x)+\frac{d}{d x}\left(3 x^{-\frac{1}{2}}\right)\\\\ =&3 x^{2}-2-\frac{3}{2 x^{\frac{3}{2}}} \end{aligned}$



2. Find $\dfrac{d y}{d x}$.
   
   (a) $y=3 x^{2}$
   
   (b) $y=\dfrac{1}{x}$
   
   (c) $y=\sqrt{x}+\dfrac{1}{\sqrt{x}}$
   
   (d) $y=x^{3}+2 x^{2}-3 x-6$
   
   (e) $y=x\left(1-x^{2}\right)^{2}$
   
   (f) $y=x^{\dfrac{3}{4}}+\dfrac{6}{x^{\dfrac{2}{3}}}$
   
   (g) $y=\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^{2}$
   
   (h) $y=\dfrac{(1-x)(3 x+2)}{\sqrt{x}}$
   
   (i) $y=\left(x-1+\dfrac{1}{x}\right)\left(x-1-\dfrac{1}{x}\right)$
   
   





$\begin{aligned} \text{(a)}\quad y &=3 x^{2}\\\\ \frac{d y}{d x}&=6 x\\\\ \text{(b)}\quad y &=\frac{1}{x}\\\\ &=x^{-1}\\\\ \frac{d y}{d x}&=-x^{-2}\\\\ &=-\frac{1}{x^{2}}\\\\ \text{(c)}\quad y&=\sqrt{x}+\frac{1}{\sqrt{x}}\\\\ &=x^{\frac{1}{2}}+x^{-\frac{1}{2}}\\\\ \frac{d y}{d x}&=\frac{1}{2} x^{-\frac{1}{2}}-\frac{1}{2} x^{-\frac{3}{2}}\\\\ &=\frac{1}{2}\left[\frac{1}{x^{\frac{1}{2}}}-\frac{1}{x^{\frac{3}{2}}}\right]\\\\ \text{(d)}\quad y&=x^{3}+2 x^{2}-3 x-6\\\\ \frac{d y}{d x}&=3 x^{2}+4 x-3\\\\ \text{(e)}\quad y&=x\left(1-x^{2}\right)^{2}\\\\ &=x\left(1-2 x^{2}+x^{4}\right)\\\\ &=x-2 x^{4}+x^{5}\\\\ \frac{d y}{d x}&=1-8 x^{3}+5 x^{4}\\\\ \text{(f)}\quad y&=x^{\frac{3}{4}}+\frac{6}{x^{\frac{2}{3}}}\\\\ &=x^{\frac{3}{4}}+6 x^{-\frac{2}{3}}\\\\ \frac{d y}{d x}&=\frac{3}{4} x^{-\frac{1}{4}}-4 x^{-\frac{5}{3}}\\\\ &=\frac{3}{4 x^{\frac{1}{4}}}-\frac{4}{x^{\frac{5}{3}}}\\\\ \text{(g)}\quad y&=\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}\\\\ &=x+2+\frac{1}{x}\\\\ &=2+x+x^{-1}\\\\ \frac{d y}{d x}&=1-x^{-2}\\\\ &=1-\frac{1}{x^{2}}\\\\ \text{(h)}\quad y&=\frac{(1-x)(3 x+2)}{\sqrt{x}}\\\\ &=\frac{2+x-3 x^{2}}{\sqrt{x}}\\\\ &=2 x^{-\frac{1}{2}}+x^{\frac{1}{2}}-3 x^{\frac{3}{2}}\\\\ \frac{d y}{d x}&=-x^{-\frac{3}{2}}+\frac{1}{2} x^{-\frac{1}{2}}-\frac{9}{2} x^{\frac{1}{2}}\\\\ &=-\frac{1}{x^{\frac{3}{2}}}+\frac{1}{2 x^{\frac{1}{2}}}-\frac{9}{2} x^{\frac{1}{2}}\\\\ \text{(i)}\quad y&=\left(x-1+\frac{1}{x}\right)\left(x-1-\frac{1}{x}\right)\\\\ &=(x-1)^{2}-\frac{1}{x^{2}}=x^{2}-2 x+1-x^{-2} \\\\ \frac{d y}{d x}&=2 x-2+2 x^{-3}\\\\ &=2 x-2+\frac{2}{x^{3}} \end{aligned}$



3. Given $f(x)=\left(x^{2}-3\right)^{2}$, find $f^{\prime}(x)$ and $f^{\prime}(-1)$.





$\begin{aligned} f(x) &=\left(x^{2}-3\right)^{2}\\\\ &=x^{4}-6 x^{2}+9\\\\ f^{\prime}(x)&=4 x^{3}-12 x \\\\ f^{\prime}(-1)&=4(-1)^{3}-12(-1)=8 \end{aligned}$



4. Given that $f(x)=4 x^{\frac{3}{2}}$, find $f^{\prime}(x)$ and $f^{\prime}(1), f^{\prime}(4), f^{\prime}\left(\dfrac{1}{9}\right)$.





$\begin{aligned} f(x)&=4 x^{\frac{3}{2}} \\\\ f^{\prime}(x)&=6 x^{\frac{1}{2}} \\\\ f^{\prime}(1)&=6(1)^{\frac{1}{2}}\\\\&=6 \\ f^{\prime}(4)&=6(4)^{\frac{1}{2}}\\\\&=12 \\ f^{\prime}\left(\frac{1}{9}\right)&=6\left(\frac{1}{9}\right)^{\frac{1}{2}}\\\\&=2 \end{aligned}$



5. Calculate the rate of change of $f: x \mapsto \sqrt[3]{x}+\dfrac{1}{\sqrt[3]{x}}$ at $x=8$.





$\begin{aligned} f(x) &=\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}\\\\ &=x^{\frac{1}{3}}+x^{-\frac{1}{3}} \\\\ f^{\prime}(x)&=\frac{1}{3} x^{-\frac{2}{3}}-\frac{1}{3} x^{-\frac{4}{3}}\\\\ &=\frac{1}{3}\left[\frac{1}{x^{\frac{2}{3}}}-\frac{1}{x^{\frac{4}{3}}}\right] \\\\ f^{\prime}(8)&=\frac{1}{3}\left[\frac{1}{8^{\frac{2}{3}}}-\frac{1}{8^{\frac{4}{3}}}\right]\\\\ &=\frac{1}{3}\left[\frac{1}{4}-\frac{1}{16}\right]\\\\ &=\frac{1}{3}\left[\frac{4-1}{16}\right]\\\\ &=\frac{1}{16} \end{aligned}$



6. Given that $A=2 r^{2}-4 r+5$, find the rate of change of $A$ with respect to $r$ when $r=3$.





$\begin{aligned} A&=2 r^{2}-4 r+5\\\\ \frac{d A}{d r}&=4 r-4\\\\&=4(r-1) \\\\ \left.\frac{d A}{d r}\right|_{r=3}&=4(3-1)\\\\&=8 \end{aligned}$



7. Given that $V=\dfrac{4}{3} r^{3}-\dfrac{3}{4} r^{2}+r-5$, find the rate of change of $V$ with respect to $r$ when $r=2$.





$\begin{aligned} V&=\frac{4}{3} r^{3}-\frac{3}{4} r^{2}+r-5 \\\\ \frac{d V}{d r}&=4 r^{2}-\frac{3}{2} r+1 \\\\ \left.\frac{d V}{d r}\right|_{r=2}&=4(2)^{2}-\frac{3}{2}(2)+1\\\\&=16-3+1\\\\&=14 \end{aligned}$



8. If a ball is thrown into the air with a velocity of $40 \mathrm{ft} / \mathrm{s}$, its height (in feet) after $t$ seconds is given by $y=40 t-16 t^{2}$. Find the velocity when $t=2$.





$y=40 t-16 t^{2}\\\\ $ Let the velocity of the ball be v. Since velocity is the rate of change of displacement with respect to time, $$\begin{aligned} &\\\\ v&=\frac{d y}{d t}\\\\&=40-32 t \\\\ \text { When } t&=2,\\\\ v&=\left.\frac{d y}{d t}\right|_{t=2}\\\\ &=40-32(2)\\\\&=-24 \mathrm{ft} / \mathrm{s}\\\\ \end{aligned}$$ After $t$ seconds, the ball is falling down with a velocity of $24 \mathrm{ft} / \mathrm{s}$.



9. The displacement (in meters) of a particle moving in a straight line is given by the equation of motion $s=\dfrac{1}{t^{2}}$, where $t$ is measured in seconds. Find the velocity of the particle at times $t=a, t=1, t=2$, and $t=3$.





$s=\dfrac{1}{t^{2}}\\\\ $ Let the velocity of the particle be $v$. Since velocity is the rate of change of displacement with respect to time, $\begin{aligned} &\\\\ v=\dfrac{d s}{d t}=-\dfrac{2}{t^{3}}\\\\ \end{aligned}$ When $t=a, v=\left.\dfrac{d s}{d t}\right|_{t=a}=-\dfrac{2}{a^{3}} \mathrm{~m} / \mathrm{s}\\\\ $ When $t=1, v=\left.\dfrac{d s}{d t}\right|_{t=1}=-\dfrac{2}{1^{3}}=-2 \mathrm{~m} / \mathrm{s}\\\\ $ When $t=2, v=\left.\dfrac{d s}{d t}\right|_{t=2}=-\dfrac{2}{2^{3}}=-0.5 \mathrm{~m} / \mathrm{s}\\\\ $ When $t=3, v=\left.\dfrac{d s}{d t}\right|_{t=3}=-\dfrac{2}{3^{3}}=-0.074 \mathrm{~m} / \mathrm{s}\\\\ $



10. The position of a stone thrown from a bridge is given by $s=10 t-16 t^{2}$ feet (below the bridge) after $t$ seconds.
    
    (a) What is the average velocity of the stone between $t_{1}=1$ and $t_{2}=5$ seconds?
    
    (b) What is the instantaneous velocity of the stone at $t=1$ second.
    
    





$s=10 t-16 t^{2}\\\\ $ When $t_{1}=1, s_{1}=10-16=-6 \mathrm{ft}\\\\ $ When $t_{2}=5, s_{2}=10(5)-16(5)^{2}=-350 \mathrm{ft}\\\\ $ $\therefore \delta s=-350-(-6)=-344 \mathrm{ft}$ and $\delta t=5-1=4 \text{seconds}\\\\ $ $\therefore$ Average velocity $=\frac{\delta s}{\delta t}=-\frac{344}{4}=-86 \mathrm{ft} / \mathrm{s}\\\\ $ $v=\frac{d s}{d t}=10-\left.32 t \Rightarrow \frac{d s}{d t}\right|_{t=1}=10-32=-22 \mathrm{ft} / \mathrm{s}\\\\ $ $\therefore$ Instantaneous velocity of the stone at $t=1$ second is $-22 \mathrm{~ft} / \mathrm{s}\\\\ $ The negative sign of the velocity means that the stone is decelerating.



11. The displacement (in metre) of a particle moving in a straight line is given by $s=t^{4}-4 t^{3}-20 t^{2}+20 t, t \geq 0$ where $t$ is measured in seconds.
    
    (a) At what time does the particle have a velocity of $20 \mathrm{~m} / \mathrm{s}$ ?
    
    (b) At what time is the acceleration 0 ?
    
    





$s=t^{4}-4 t^{3}-20 t^{2}+20 t\\\\ $ Let the velocity of and acceleration of the particle be $v$ and $a$ respectively. Since velocity is the rate of change of displacement and acceleration is the rate of change of velocity with respect to time, $\begin{aligned} v&=\frac{d s}{d t}\\\\ &=4 t^{3}-12 t^{2}-40 t+20\\\\ &=4\left(t^{3}-3 t^{2}-10 t+5\right) \mathrm{m} / \mathrm{s} \\\\ a&=\frac{d v}{d t}\\\\ &=12 t^{2}-24 t-40\\\\ &=4\left(3 t^{2}-6 t-10\right) \mathrm{m} / \mathrm{s}^{2} \\\\ \end{aligned}$ $\begin{aligned} \text { When } v=20 \mathrm{~m} / \mathrm{s},\quad\quad & \\\\ 4\left(t^{3}-3 t^{2}-10 t+5\right)&=20 \\\\ \therefore t^{3}-3 t^{2}-10 t&=0 \\\\ t\left(t^{2}-3 t-10\right)&=0\\\\ t(t-5)(t+2)&=0 \\\\ \therefore\quad t=0 \text { or } t&=5(t \geq 0) \\\\ \text { When } a&=0,\\\\ 4\left(3 t^{2}-6 t-10\right)&=0 \\\\ \therefore 3 t^{2}-6 t-10&=0 \\\\ t^{2}-2 t&=\frac{10}{3} \\\\ t^{2}-2 t+1&=\frac{13}{3} \\\\ \therefore(t-1)^{2}&=\frac{13}{3} \\\\ t-1&=2.082\\\\ t&=3.082 \text { seconds } \end{aligned}$



12. When a marble is moving in a groove, the distance $s$ centimetres from one end at time $t$ second is given by $s=5 t-t^{2}$. Find the speed of marble at $t=2$. Find $t$ when the speed of marble is zero.





$s=5 t-t^{2}\\\\ $ Let the speed of the marble be $v$. $\begin{aligned} &\\\\ v&=\frac{d s}{d t}\\\\ &=5-2 t\\\\ \text { When } t&=2,\\\\ \left.\frac{d s}{d t}\right|_{t=2}&=5-2(2)\\\\ &=1 \mathrm{~cm} / \mathrm{s} \\\\ \text { When } v&=0,\\\\ 5-2 t&=0 \\\\ \therefore t&=2.5 \text { seconds } \end{aligned}$

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