MATH 1830

Unit 1 Limits

1.4 Derivatives: The Power Rule

Basic Differentiation Properties

Three Equivalent Terms:

If $y\; = \;\;f\left( x \right),\;$ you can use any of these to represent the derivative $y' = f'\left( x \right) = \frac{{dy}}{{dx}}$

THE POWER RULE: If $f\left( x \right) = {x^n}$ then $f'\left( x \right) = n{x^{n - 1}}$



1.4A Video

Using the Power Rule, find the indicated derivative:

  1. $g\left( x \right) = {x^4}$

    $g'\left( x \right) = 4{x^3}$

  2. $y = 2{x^3}$

    $y' = 2\times3{x^{ 2}}$

    $y' = 6{x^{ 2}}$

  3. $\frac{d}{{dx}}\left( 5 \right)$

    $\frac{d}{{dx}}\left( 5 \right) = \frac{d}{{dx}}\left( {5{x^0}} \right) = 5\left( 0 \right){x^{ - 1}} =$

    $0$

  4. $y = \;\frac{1}{{{x^7}}}$

    $y=x^{-7}$

    $y' = - 7{x^{ - 8}} = \frac{{ - 7}}{{{x^8}}}$

  5. $y = \;\frac{{{x^4}}}{{16}}$

    $\frac{{dy}}{{dx}} = \frac{{4{x^3}}}{{16}} = \frac{{{x^3}}}{4}$



    6. 1.4A Lecture

    Using the Power Rule, find the indicated derivative:


  6. $y\; = 8 + 3t - 5{t^3}$

    $\frac{{dy}}{{dt}} = 0 + 3 - 15{t^2} = 3 - 15{t^2}$

  7. $g\left( x \right) = 6{x^{ - 5}} - \;\;5{x^{ - 4}}$

    $g'\left( x \right) = - 30{x^{ - 6}} + 20{x^{ - 5}} =$

    $ \frac{{ - 30}}{{{x^6}}} + \frac{{20}}{{{x^5}}}$

  8. $\frac{d}{{dx}}\left( {\frac{{2{x^3}}}{{5}}\; - \;\;\frac{2}{{3{x^4}}}} \right)$

    $\frac{d}{{dx}}\left( {\frac{2}{{5}}{x^3} - \frac{2}{3}{x^{ - 4}}} \right)$

    $\frac{d}{{dx}} = \frac{{6}}{{5}}{x^2} + \frac{8}{3}{x^{ - 5}} = $

    $\frac{{6{x^2}}}{5} + \frac{8}{{3{x^5}}}$

  9. $q(x)=\frac{{{x}^{3}}-x+2}{x}$

    $q(x)=\frac{x^3}{x}-\frac{x}{x}+\frac{2}{x}$

    $q(x)=x^2-1+2{x^{-1}}$

    ${q}'(x)=2x-2{{x}^{-2}}$

    ${q}'(x)=2x-\frac{2}{{{x}^{2}}}$

    10. 1.4A Group Work

    Using the Power Rule, find the indicated derivative:


  10. $p(a)=3{{a}^{4}}-2{{a}^{3}}+7{{a}^{2}}-a+12$

    ${p}'(a)=12{{a}^{3}}-6{{a}^{2}}+14a-1$

  11. $\frac{d}{dx}\left( -5{{x}^{4}}-6{{x}^{2}}-\frac{2}{{{x}^{3}}} \right)$

    $\frac{d}{dx}(-5{{x}^{4}}-6{{x}^{2}}-2{{x}^{-3}})$

    $\frac{d}{dx}=-20{{x}^{3}}-12x+6{{x}^{-4}}$

    $\frac{d}{dx}=-20{{x}^{3}}-12x+\frac{6}{{{x}^{4}}}$

  12. A company’s total sales (in millions of dollars) t months from now are given by $S\left( t \right) = 0.032{t^4}\;\; + 0.5{t^3}\; + 2.8{t^2}\; + 9t - 4.$

    1. Find $S'\left( t \right).$

      $S'\left( t \right) = 0.128{t^3} + 1.5{t^2} + 5.6t + 9$

    2. Find $S\left( 4 \right)$ and $S'\left( 4 \right).$ Write a brief verbal interpretation of these results.

      $S(4) = 0.032{(4)^4} + 0.5{(4)^3} + 2.8{(4)^2} + 9(4) - 4 = 116.99$

      $S'(4) = 0.128{(4)^3} + 1.5{(4)^2} + 5.6(4) + 9 = 63.59$

      Total sales after 4 months are \$116.99 million. After 4 months, the sales are increasing at a rate of $63.59 million per month.

    3. Find $S\left( 8 \right)$ and $S'\left( 8 \right).$ Write a brief verbal interpretation of these results.

      $S(8) = 0.032{(8)^4} + 0.5{(8)^3} + 2.8{(8)^2} + 9(8) - 4 = 634.27$

      $S'(8) = 0.128{(8)^3} + 1.5{(8)^2} + 5.6(8) + 9 = 215.34$

      Total sales after 8 months are \$634.27 million. After 8 months, the sales are increasing at a rate of $215.34 million per month.

    1. Find the slope of the tangent line to the curve $p(x)=3{{x}^{4}}-2{{x}^{3}}+7{{x}^{2}}-x+12$ at the point where $x=-1$.

      ${p}'(x)=12{{x}^{3}}-6{{x}^{2}}+14x-1$

      ${m_{\tan }}={p}'\left( -1 \right)=12{{(-1)}^{3}}-6{{(-1)}^{2}}+14(-1)-1=-33$

      At $x=-1, {m_{\tan }}=-33$

    2. Find the equation of the tangent line at $x=-1$.

      $(-1,25)$

      $y-25=-33(x+1)$

      $y-25=-33x-33$

      $y=-33x-8$

      Active Calculus by Matthew Boelkins is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. Based on a work at http://scholarworks.gvsu.edu/books/10/.



    14. 1.4A Additional Practice


  14. $f(x)={{x}^{7}}$

    ${f}'(x)=7{{x}^{6}}$

  15. $h(z)=\pi $

    ${h}'(z)=0$

  16. $y=\frac{1}{{{x}^{9}}}$

    $y=x^{-9}$

    ${y}'=-9{{x}^{-10}}=\frac{-9}{{{x}^{10}}}$

  17. $\frac{d}{dx}\left( \frac{{{x}^{6}}}{36} \right)$

    $\frac{d}{dx}=\frac{6{{x}^{5}}}{36}=\frac{{{x}^{5}}}{6}$

  18. $f(x)=\frac{2{{x}^{5}}-3{{x}^{3}}+x}{{{x}^{2}}}$

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

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

    $f(x)=2{{x}^{3}}-3x+{{x}^{-1}}$

    ${f}'(x)=6{{x}^{2}}-3-1{{x}^{-2}}$

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

1.4B Derivatives: The Power Rule


1.4B Video

  1. $\frac{d}{{du}}\left( {7{u^{2/3}}\;\; + \;\;\;4{u^{ - 3/5}}} \right)$

    $\frac{d}{{du}} = \frac{{14}}{3}{u^{ - 1/3}} - \frac{{12}}{5}{u^{ - 8/5}}$

  2. $H'\left( w \right)$ if $H\left( w \right) = \;\frac{5}{{{w^6}}}\; - \;\;7\sqrt w$

    $H(w)= 5{w^{ - 6}} - 7{w^{1/2}}$

    $H'(w) = - 30{w^{ - 7}} - \frac{7}{2}{w^{ - 1/2}} =$

    $ \frac{{ - 30}}{{{w^7}}} - \frac{7}{{2\sqrt w }}$



    3. 1.4B Lecture

    Determine the derivative of each of the following functions. State your answer using full and proper notation, labeling the derivative with its name. For example, if you are given a function $h(z),$ you should write $h'(z)\ $or $\frac{dh}{dz}$ as part of your response.

  3. $f(x)=\sqrt[7]{{{x}^{4}}}$$={x}^{4/7}$

    ${f}'(x)=\frac{4}{7}{{x}^{-3/7}}=\frac{4}{7\sqrt[7]{{{x}^{3}}}}$

  4. $h(x)=\frac{5}{\sqrt{x}}$$=5{x}^{-1/2}$

    ${h}'(x)=\frac{-5}{2}{{x}^{-3/2}}=\frac{-5}{2\sqrt{{{x}^{3}}}}$

    ${h}'(x)=\frac{-5}{2x\sqrt{x}}$

    5.  

    1.4B Group Work

    Determine the derivative of each of the following functions. State your answer using full and proper notation, labeling the derivative with its name. For example, if you are given a function $h(z),$ you should write $h'(z)\ $or $\frac{dh}{dz}$ as part of your response.

  5. $y=\frac{{{x}^{6}}}{12}-3\,\sqrt[3]{x}$

    $y=\frac{{{x}^{6}}}{12}-3{{x}^{1/3}}$

    ${y}'=\frac{6{{x}^{5}}}{12}-3(\frac{1}{3}){{x}^{-2/3}}$

    ${y}'=\frac{{{x}^{5}}}{2}-{{x}^{-2/3}}$

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

  6. $\frac{d}{dx}\left( -5\sqrt[4]{{{x}^{3}}}-6\sqrt[3]{{{x}^{2}}}-\frac{2}{{{x}^{3}}} \right)$

    $\frac{d}{dx}\left( -5{{x}^{3/4}}-6{{x}^{2/3}}-2{{x}^{-3}} \right)$

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

    $\frac{d}{dx}=\frac{-15}{4}{{x}^{-1/4}}-4{{x}^{-1/3}}+6{{x}^{-4}}$

    $\frac{d}{dx}=\frac{-15}{4\sqrt[4]{x}}-\frac{4}{\sqrt[3]{x}}+\frac{6}{{{x}^{4}}}$

  7. The population of a bacteria colony is modeled by the function $p(t)=200+20t-{{t}^{2}}$, where t is time in hours, $t\ge 0$, and p is the number of bacteria, in thousands.

    ${p}'(t)=20-2t$

    1. Determine the growth rate of the bacteria population at each of the following times.
      1. 3 hours

        ${p}'(3)=20-2(3)=20-6=14$

        At 3 hours, the bacteria population is increasing at a rate of 14 thousand bacteria per hour.

      2. 8 hours

        ${p}'(8)=20-2(8)=20-16=4$

        At 8 hours, the bacteria population is increasing at a rate of 4 thousand bacteria per hour.

      3. 13 hours

        ${p}'(13)=20-2(13)=20-26=-6$

        At 13 hours, the bacteria population is decreasing at a rate of 6 thousand bacteria per hour.

      4. 18 hours

        ${p}'(18)=20-2(18)=20-36=-16$

        At 18 hours, the bacteria population is decreasing at a rate of 16 thousand bacteria per hour.

    2. What are the implications of the growth rates in part a?

      As time increases, growth rate slows and eventually the bacteria start to die.

    3. Determine the equation of the tangent to p(t) at the point t=8.

      $(8,296)$

      ${m_{\tan }}=4$ from problem a part ii above

      $y-296=4(x-8)$

      $y-296=4x-32$

      $y=4x+264$

    4. When does the bacteria population stop growing?

      The population stops growing when ${p}'(t)=0$

      $20-2t=0$

      $10=t$

      The population stops growing at 10 hours.

    5. What is the population at this time?

      $p(10)=200+20(10)-(10)^2=300,000$

      The population at 10 hours is 300,000 bacteria.

    8.  

    1.4B Additional Practice

  8. $\frac{d}{dx}\left( -\frac{1}{{{x}^{4}}}\ +\ 9{{x}^{5}} \right)$

    $\frac{d}{dx}(-{{x}^{-4}}+9{{x}^{5}})$

    $\frac{d}{dx}=4{{x}^{-5}}+45{{x}^{4}}$

    $\frac{d}{dx}=\frac{4}{{{x}^{5}}}+45{{x}^{4}}$

  9. $p(a)=-3{{a}^{-5}}+{{a}^{-3}}+7a-\sqrt{a}+12$

    $p(a)=-3{{a}^{-5}}+{{a}^{-3}}+7a-{{a}^{1/2}}+12$

    ${p}'(a)=15{{a}^{-6}}-3{{a}^{-4}}+7-\frac{1}{2}{{a}^{-1/2}}$

    ${p}'(a)=\frac{15}{{{a}^{6}}}-\frac{3}{{{a}^{4}}}+7-\frac{1}{2\sqrt{a}}$

  10. $q(x)=\frac{-8{{x}^{3}}+5x-6}{{{x}^{4}}}$

    $q(x)=-8{{x}^{-1}}+5{{x}^{-3}}-6{{x}^{-4}}$

    ${q}'(x)=8{{x}^{-2}}-15{{x}^{-4}}+24{{x}^{-5}}$

    ${q}'(x)=\frac{8}{{{x}^{2}}}-\frac{15}{{{x}^{4}}}+\frac{24}{{{x}^{5}}}$

  11. Find the equation of the tangent line to the graph of $g(x)=-\frac{5}{\sqrt{x}}$ at the point $x=4$.

    $g(4)=\frac{-5}{\sqrt4}$

    $g(4)=\frac{-5}2$

    Use point $(4,\frac{-5}2)$

    $g(x)=-5x^{-1/2}$

    ${g}'\left( x \right)=\frac{5}{2}{{x}^{-3/2}}$

    ${m_{\tan }}={g}'\left( 4 \right)=\frac{5}{2}{{\left( 4 \right)}^{-3/2}}=\frac{5}{16}$

    $y+\frac{5}{2}=\frac{5}{16}\left( x-4 \right)$

    $y+\frac{5}{2}=\frac{5}{16}x-\frac54$

    $y=\frac{5}{16}x-\frac{15}{4}$