MATH 1830

2.4 Derivatives of Quotients

Pre-Class:

• Complete 2.3 Homework assignment: check and correct.
• Take notes on the videos and readings (use the space below).
• Work and check problem #1 in the 2.4 NOTES section.
• Complete the 2.4 Pre-Class Quiz.

Introduction

1. The cost of manufacturing x MP3 players per day is represented by the function $$C(x) = 0.01{x^2} + 42x + 300\quad 0 \le x \le 300.$$

   1. Determine the average cost function.

      $\bar C(x) = \frac{{0.01{x^2} + 42x + 300}}{x}$

   2. Determine the marginal average cost function. What did you have to do to the average cost function in order to find the marginal average cost function?

      The marginal average cost is the derivative of the average cost.

      Must rewrite the function to use the power rule.

      $\bar C(x) = 0.01x + 42 + \frac{{300}}{x}$

      $\bar C(x) = 0.01x + 42 + 300{x^{ - 1}}$

      $\bar C'(x) = 0.01 - 300{x^{ - 2}}$

      $\bar C'(x) = 0.01 - \frac{{300}}{{{x^2}}}$

2. Suppose the function $V(t) = \frac{{50,000 + 6t}}{{1 + 0.4t}}$ represents the value, in dollars, of a new car t years after it is purchased. Determine the rate of change in the value of the car.

   We cannot rewrite $V(t)$ to use the power rule as we did for the question above. We need another new rule.

   ---

Notes

Derivatives of Quotients

Rewriting a Function as a Quotient

The Quotient Rule

If $y = \;\frac{f(x)}{g(x)},$

then $y' = \;\frac{{f'(x)\;g(x)\; - \;f(x)\;g'(x)\;}}{{[{g(x)}}]^2}.$

Two Methods for Finding the Derivative:

Find the derivative two different ways.

1. Simplify first and use the power rule.
2. Use the quotient rule.




1. $r\left( x \right) = \;\frac{{{x^5}+4}}{{{x^2}}}$

   1. Simplifying and Using Power Rule

      $r(x) = \frac{{{x^5}}}{{{x^2}}} + \frac{4}{{{x^2}}}$

      $r(x) = {x^3} + 4{x^{ - 2}}$

      ${r'}(x) = 3{x^2} - 8{x^{ - 3}}$

      ${r'}(x) = 3{x^2} - \frac{8}{{{x^3}}}$

   2. Using Quotient Rule

      $f = {x^5} + 4$

      $f' = 5{x^4}$

      $g = {x^2}$

      $g' = 2x$

      $r'(x) = \frac{{5{x^4}({x^2}) - 2x({x^5} + 4)}}{{({x^2})^2}}$

      $r'(x) = \frac{{5{x^6} - 2{x^6} - 8x}}{{{x^4}}}$

      $r'(x) = \frac{{3{x^6} - 8x}}{{{x^4}}}$

      $r'(x) = \frac{{x(3{x^5} - 8)}}{{{x^4}}}$

      $r'(x) = \frac{{3{x^5} - 8}}{{{x^3}}}$

      $r'(x) = 3{x^2} - \frac{8}{{{x^3}}}$ (which is the same answer as part a)

   Find the Derivative of each Function using the Quotient Rule.

2. $b\left( x \right) = \;\frac{{4x}}{{3x + 8}}$

   $f =4x$

   $f' =4$

   $g =3x+8$

   $g' = 3$

   $b'(x) = \frac{{4(3x + 8) - 3(4x)}}{{{{\left( {3x + 8} \right)}^2}}}$

   $b'(x) = \frac{{12x + 32 - 12x}}{{{{\left( {3x + 8} \right)}^2}}}$

   $b'(x) = \frac{{32}}{{{{\left( {3x + 8} \right)}^2}}}$

3. $c\left( x \right) = \;\frac{{{x^2}\; - 9}}{{{x^2}\; + 1}}$

   $f ={x^2} - 9$

   $f' =2x$

   $g ={x^2} + 1$

   $g' = 2x$

   $c'(x) = \frac{{2x({x^2} + 1) - 2x({x^2} - 9)}}{{{{\left( {{x^2} + 1} \right)}^2}}}$

   $c'(x) = \frac{{2{x^3} + 2x - 2{x^3} + 18x}}{{{{\left( {{x^2} + 1} \right)}^2}}}$

   $c'(x) = \frac{{20x}}{{{{\left( {{x^2} + 1} \right)}^2}}}$

4. $h\left( x \right) = \;\frac{{1 + {e^x}}}{{1 - {e^x}}}$

   $f =1 + {e^x}$

   $f' = {e^x}$

   $g = 1 - {e^x}$

   $g' = - {e^x}$

   $h'(x) = \frac{{{e^x}(1 - {e^x}) - \left( { - {e^x}} \right)(1 + {e^x})}}{{{{\left( {1 - {e^x}} \right)}^2}}}$

   $h'(x) = \frac{{{e^x}(1 - {e^x}) + {e^x}(1 + {e^x})}}{{{{\left( {1 - {e^x}} \right)}^2}}}$

   $h'(x) = \frac{{{e^x} - {e^{2x}} + {e^x} + {e^{2x}}}}{{{{\left( {1 - {e^x}} \right)}^2}}}$

   $h'(x) = \frac{{2{e^x}}}{{{{\left( {1 - {e^x}} \right)}^2}}}$

5. $j\left( x \right) = \;\frac{{3x}}{{4 + \ln x}}$

   $f =3x$

   $f' = 3$

   $g =4 + \ln x$

   $g' = \frac{1}{x}$

   $j'(x) = \frac{{3(4 + \ln x) - \frac{1}{x}(3x)}}{{{{\left( {4 + \ln x} \right)}^2}}}$

   $j'(x) = \frac{{12 + 3\ln x - 3}}{{{{\left( {4 + \ln x} \right)}^2}}}$

   $j'(x) = \frac{{9 + 3\ln x}}{{{{\left( {4 + \ln x} \right)}^2}}}$

6. Find $\frac{{dy}}{{dw}}$ for $y = \;\frac{{2{w^4}\; - \;{w^3}}}{{6w - 1}}$

   $f =2{w^4} - {w^3}$

   $f' = 8{w^3} - 3{w^2}$

   $g = 6w - 1$

   $g' = 6$

   $\frac{{dy}}{{dw}} = \frac{{(8{w^3} - 3{w^2})(6w - 1) - 6(2{w^4} - {w^3})}}{{{{\left( {6w - 1} \right)}^2}}}$

   $\frac{{dy}}{{dw}} = \frac{{48{w^4} - 8{w^3} - 18{w^3} + 3{w^2} - 12{w^4} + 6{w^3}}}{{{{\left( {6w - 1} \right)}^2}}}$

   $\frac{{dy}}{{dw}} = \frac{{36{w^4} - 20{w^3} + 3{w^2}}}{{{{\left( {6w - 1} \right)}^2}}}$

7. Explain how $f'(x)$ can be found without using the quotient rule: $f\left( x \right) = \;\frac{4}{{{x^3}}}.$

   Rewrite the quotient as a product and then use the product rule.

   $f(x) = 4{x^{ - 3}}$

   $f'(x) = - 12{x^{ - 4}}$

   $f'(x) = \frac{{ - 12}}{{{x^4}}}$

   Tangent Lines

8. $h\left( x \right) = \;\frac{{3x - 7}}{{2x - 1}}$

   1. Find $h'\left( x \right).$

      $f =3x - 7$

      $f' = 3$

      $g = 2x - 1$

      $g' = 2$

      $h'(x) = \frac{{3(2x - 1) - 2(3x - 7)}}{{{{\left( {2x - 1} \right)}^2}}}$

      $h'(x) = \frac{{6x - 3 - 6x + 14}}{{{{\left( {2x - 1} \right)}^2}}}$

      $h'(x) = \frac{{11}}{{{{\left( {2x - 1} \right)}^2}}}$

   2. Find the equation of the line tangent to the graph of $h$ at $x\; = \;2$.

      Point $\quad h(2) = \frac{{3\cdot2 - 7}}{{2\cdot2 - 1}} = \frac{{ - 1}}{3}$

      $(2, - \frac{1}{3})$

      Slope $\quad {m_{tan}} = f'(2) = \frac{{11}}{{{{\left( {2\cdot2 - 1} \right)}^2}}} = \frac{{11}}{{{3^2}}} = \frac{{11}}{9}$

      ${m_{tan}} = \frac{{11}}{9}$

      Equation of Line: $y + \frac{1}{3} = \frac{{11}}{9}(x - 2)$

      $y + \frac{1}{3} = \frac{{11}}{9}x - \frac{{22}}{9}$

      $y = \frac{{11}}{9}x - \frac{{25}}{9}$

   3. Find the values of x where h’(x) = 0.

      $\frac{{11}}{{{{(2x - 1)}^2}}} = 0$

      $\frac{{11}}{{{{(2x - 1)}^2}}} = \frac{0}{1}$

      Cross Multiply

      $0 \ne 11$ There is no value of x where $h'(x) =0.$

   Derivatives with Radicals

9. Find y’ for $y = \;\frac{{6\sqrt[3]{x}}}{{2{x^2}\; - 5x + 1}}.$

   $f(x)=\frac{{6{x^{1/3}}}}{{2{x^2} - 5x + 1}}$

   $f =6{x^{1/3}}$

   $f' = \frac{1}{3}\cdot 6{x^{ - 2/3}}$

   $f' = 2{x^{ - 2/3}}$

   $g = 2{x^2} - 5x + 1$

   
   

   $g' = 4x - 5$

   
   

   $y' = \frac{{2{x^{ - 2/3}}(2{x^2} - 5x + 1) - (4x - 5)(6{x^{1/3}})}}{{{{\left( {2{x^2} - 5x + 1} \right)}^2}}}$

   $y' = \frac{{4{x^{4/3}} - 10{x^{1/3}} + 2{x^{ - 2/3}} - 24{x^{4/3}} + 30{x^{1/3}}}}{{{{\left( {2{x^2} - 5x + 1} \right)}^2}}}$

   $y' = \frac{{ - 20{x^{4/3}} + 20{x^{1/3}} + 2{x^{ - 2/3}}}}{{{{\left( {2{x^2} - 5x + 1} \right)}^2}}}$

   Multiply every term by ${{x}^{2/3}}.$

   $y' = \frac{{ - 20{x^2} + 20x + 2}}{{\sqrt[3]{{{x^2}}}{{(2{x^2} - 5x + 1)}^2}}}$

10. Find $\frac{{dy}}{{dx}}$ for $y = \;\frac{{\;2{x^2}\; - 2x + 3\;}}{{\sqrt[4]{x}}}.$

    $f =2{x^2} - 2x + 3$

    $f' = 4x - 2$

    $g = {x^{1/4}}$

    $g' = \frac{1}{4}{x^{ - 3/4}}$

    $\frac{{dy}}{{dx}} = \frac{{(4x - 2)({x^{1/4}}) - \frac{1}{4}{x^{ - 3/4}}(2{x^2} - 2x + 3)}}{{{{\left( {{x^{1/4}}} \right)}^2}}}$

    $\frac{{dy}}{{dx}} = \frac{{4{x^{5/4}} - 2{x^{1/4}} - \frac{1}{2}{x^{5/4}} + \frac{1}{2}{x^{1/4}} - \frac{3}{4}{x^{ - 3/4}}}}{{{x^{1/2}}}}$

    $\frac{{dy}}{{dx}} = \frac{{\left( {\frac{7}{2}{x^{5/4}} - \frac{3}{2}{x^{1/4}} - \frac{3}{4}{x^{ - 3/4}}} \right)}}{{({x^{1/2}})}}$

    $\frac{{dy}}{{dx}} = \frac{{\left( {\frac{7}{2}{x^{5/4}} - \frac{3}{2}{x^{1/4}} - \frac{3}{4}{x^{ - 3/4}}} \right)\cdot 4}}{{({x^{1/2}})\cdot 4}}$

    $\frac{{dy}}{{dx}} = \frac{{14{x^{5/4}} - 6{x^{1/4}} - 3{x^{ - 3/4}}}}{{(4{x^{1/2}})}}$

    $\frac{dy}{dx}=\frac{14{{x}^{5/4}}-6{{x}^{1/4}}-\frac{3}{{{x}^{{}^{3}/{}_{4}}}}}{4{{x}^{1/2}}}$

    Multiply every term by ${{x}^{3/4}}.$

    $\frac{dy}{dx}=\frac{14{{x}^{8/4}}-6{{x}^{4/4}}-3}{(4{{x}^{1/2}}){{x}^{3/4}}}$

    $\frac{dy}{dx}=\frac{14{{x}^{2}}-6x-3}{4{{x}^{5/4}}}$

    Applications

11. A cable company has installed a new television system in a city. The total number N (in thousands) of subscribers t months after the installation of the system is given by $N\left( t \right) = \;\frac{{178t}}{{t + 5}}.$

    1. Find $N'(t).$

       $f =178t$

       $f' = 178$

       $g = t + 5$

       $g' = 1$

       $N'(t) = \frac{{178(t + 5) - 1(178t)}}{{{{\left( {t + 5} \right)}^2}}}$

       $N'(t) = \frac{{178t + 890 - 178t}}{{{{\left( {t + 5} \right)}^2}}}$

       $N'(t) = \frac{{890}}{{{{\left( {t + 5} \right)}^2}}}$

    2. Find $N\left( {12} \right)$ and $N'\left( {12} \right)$. Write a brief interpretation of these results.

       $N(12) = \frac{{178(12)}}{{12 + 5}} = \frac{{2136}}{{17}} = 125.647$

       $N'(12) = \frac{{890}}{{{{\left( {12 + 5} \right)}^2}}} = \frac{{890}}{{{{17}^2}}} = \frac{{890}}{{289}} = 3.0796$

       At 12 months the cable company has 125,647 subscribers and that number is increasing at a rate of 3080 subscribers per month.

    3. Use the results above to estimate the total number of subscribers after 13 months.

       $N(13)\approx N(12)+N'(12)$

       125,647 + 3080 = 128,727

       Based on the number of subscribers after 12 months, there will be approximately 128,727 subscribers after 13 months.

12. According to economic theory, the supply x of a quantity in a free market increases as the price p increases. Suppose the number x of baseball gloves a retail chain is willing to sell per week at a price of \$p is given by $$x = \;\frac{{100p}}{{0.1p + 1}} \quad\ 30.00 \le p \le 190.00.$$

1. Find $\frac{{dx}}{{dp}}.$

   $f =100p$

   $f' = 100$

   $g =0.1p + 1$

   $g' = 0.1$

   $\frac{{dx}}{{dp}} = \frac{{100(0.1p + 1) - 0.1(100p)}}{{{{\left( {0.1p + 1} \right)}^2}}}$

   $\frac{{dx}}{{dp}} = \frac{{10p + 100 - 10p}}{{{{\left( {0.1p + 1} \right)}^2}}}$

   $\frac{{dx}}{{dp}} = \frac{{100}}{{{{\left( {0.1p + 1} \right)}^2}}}$

2. Find the supply and the instantaneous rate of change (IRC) of supply with respect to price when the price is \$40. Write a brief verbal interpretation of these results.

   Supply when $p=\$40\quad\quad\quad x(40) = \frac{{100(40)}}{{0.1(40) + 1}} = \frac{{4000}}{5} = 800\;gloves$

   IRC when $p=\$40\quad\quad\quad \frac{{dx}}{{dp}}(40)=\frac{{100}}{{{{\left({0.1\cdot40+1}\right)}^2}}} = \frac{{100}}{{25}} = 4\;gloves$

   When the price of the baseball gloves is $40, the retail chain can sell 800 gloves per week, and the number of gloves sold is increasing at a rate of 4 gloves per week.

3. Use the results above to estimate the supply if the price is increased to \$41.

   800+4=804

   If the price increased to $41, 804 gloves would be sold each week.