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) $x^{3}-2 x+\dfrac{3}{\sqrt{x}}$
   
   
   


solution
$\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) $y=\left(x-1+\dfrac{1}{x}\right)\left(x-1-\dfrac{1}{x}\right)$
   
   
   



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

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


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

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

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

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


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

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


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

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 $r=2$.
   


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

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 $t=2$.
   

solution
$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 $t=3$.
   


solution
$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.
    
    
    


solution
$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 ?
    
    
    


solution
$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.
    


solution
$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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