MATH 1830

Unit 2 Derivatives

2.5 The Chain Rule

Pre-Class 2.5A:

2.4 HW

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2.5A Quiz

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Pre-Class 2.5B:

2.5A HW

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2.5B Quiz

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Introduction

  1. The gas tank of a parked pickup truck develops a leak. The amount of gas, in liters, remaining in the tank after t hours is represented by the function $V(t) = 90{\left( {1 - \frac{t}{{18}}} \right)^2}\quad 0 \le t \le 18$. How fast is the gas leaking from the tank after 12 hours?

    $f(x) = 90(1-\frac{t}{9}+\frac{t^2}{324})$

    $f(x) = 90 - 10t + \frac{5}{{18}}{t^2}$

    $f'(x) = - 10 + \frac{5}{9}t$

    $f'(12) = - 3.33$

    The gas is leaking from the tank at a rate of 3.33 liters per hour after 12 hours.

  2. Andrew and David are training to run a marathon. They both go on a run on Sunday mornings at precisely 7 A.M. Andrew’s house is 22 km south of David’s. One Sunday morning, Andrew leaves his house and runs west at 7 km/hr. The distance between the two runners can be modeled by the function $$s(t) = \sqrt {130{t^2} - 396t + 484},$$ where s is in kilometers and t is in hours. Determine the rate at which the distance between the two runners is changing.

    $s(t)={{\left( 130{{t}^{2}}-396t+484 \right)}^{{}^{1}/{}_{2}}}$

    $s'(t)=\frac{1}{2}{{\left( 130{{t}^{2}}-396t+484 \right)}^{{}^{-1}/{}_{2}}}(260t-396)$

    $s'(t)=\frac{260t-396}{2\sqrt{130{{t}^{2}}-396t+484}}$

Notes

General Derivative Rules Using the Chain Rule

$ \frac d{dx}\left[f(x)\right]^n=n\left[f(x)\right]^{n-1}\cdot f'(x)$

$\frac d{dx}\ln\left[f(x)\right]=\frac1{f(x)}\cdot f'(x)$

$\frac d{dx}e^{f(x)}=e^{f(x)}\cdot f'\left(x\right)​$

Fill in the blank with an expression that will make the indicated equation valid. Then simplify.

  1. $\frac{d}{{dx}}\;{\left( {3 - 7x} \right)^6} = 6{\left( {3 - 7x} \right)^5}\;$______ $( - 7)$

  2. $\frac{d}{{dx}}\;{e^{5x - 3}} = \;{e^{5x - 3}}$_______ $(5)$

  3. $\frac{d}{{dx\;}}\;ln\left( {{x^2} - {x^4}} \right) = \;\frac{1}{{{x^2} - {x^4}}}$______ $(2x - 4{x^3})$

    Find $f'\left( x \right)$ and simplify.

  4. $f\left( x \right) = \;{\left( {8{x^2}\; - 7} \right)^5}$

    $f'(x) = 5{\left( {8{x^2} - 7} \right)^4}(16x)$

    $f'(x) = 80x{\left( {8{x^2} - 7} \right)^4}$

  5. $f\left( x \right) = \;{e^{3{x^2}\; + 2x + 5}}$

    $f'(x) = {e^{3{x^2} + 2x + 5}}(6x + 2)$

  6. $f\left( x \right) = 2\ln \left( {9{x^2}\; - 5x + 21} \right)$

    $f'(x) = \frac{2}{{9{x^2} - 5x + 21}}\cdot(18x - 5)$

    $f'(x) = \frac{{36x - 10}}{{9{x^2} - 5x + 21}}$

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

    $f'(x) = 7{\left( {4x - 5\ln x} \right)^6}(4 - \frac{5}{x})$

    $f'(x) = 7(4 - \frac{5}{x}){\left( {4x - 5\ln x} \right)^6}$

    Horizontal Tangents

    Finding the Equation of the Tangent Line, at $x=a$:
    1. Find the y value by calculating $f(a)$: $\left(a, f(a)\right).$
    2. Find the slope of the tangent line by calculating $f'(a)$: ${m_{tan}}=f'(a).$
    3. Equation of the tangent line: $y-f(a)=f'(a)\left(x-a\right).$
    Finding the Value(s) where the Tangent Line is Horizontal:
    1. Set $f'(x)=0.$
    2. Solve for $x.$
    3. Verify that each $x$ is in the domain of $f(x)$ and $f'(x).$

    Find $f'(x)$ and simplify. Then find the equation of the tangent line to the graph of $f(x)$ at the given value of $x$. Find the values of $x$ where the tangent line is horizontal.

  8. $f\left( x \right) = \;{\left( {3x + 13} \right)^{1/2}}\quad \quad$ at $x = 4$

    $f'(x) = \frac{1}{2}{\left( {3x + 13} \right)^{ - 1/2}}(3)$

    $f'(x) = \frac{3}{{2\sqrt {3x + 13} }}$

    Slope: $\quad {m_{tan}} = f'(4) = \frac{3}{{2\sqrt {3\cdot4 + 13} }}$

    $\quad\quad{m_{tan}} = \frac{3}{{2\sqrt {12 + 13} }}$

    $\quad\quad{m_{tan}} = \frac{3}{{2\sqrt {25} }}$

    $\quad\quad{m_{tan}} = \frac{3}{{2\cdot5}} = \frac{3}{{10}}$

    Point: $f(4) = {\left( {3\cdot 4 + 13} \right)^{1/2}}$

    $\quad\quad f(4) = {\left( {12 + 13} \right)^{1/2}}$

    $\quad\quad f(4) = \sqrt {25} = 5$

    $\quad\quad (4,5)$

    $y - 5 = \frac{3}{{10}}(x - 4)$

    $y - 5 = \frac{3}{{10}}x - \frac{{12}}{{10}}$

    $y = \frac{3}{{10}}x - \frac{{6}}{{5}} + 5$

    $y = \frac{3}{{10}}x + \frac{{19}}{5}$

    Horizontal Tangent

    Horizontal Tangents are found where $f'(x) = 0.$

    $\frac{3}{{2\sqrt {3x + 13} }} = 0$

    $\frac{3}{{2\sqrt {3x + 13} }} = \frac{0}{1}$

    Cross multiply.

    $3 \ne 0$

    No Horizontal Tangents: There are no values of x where $f'(x) = 0.$

  9. $f\left( x \right) = \;3{e^{2{x^2}\; + 5x - 4}}\quad \quad x = 0$

    $f'(x) = 3{e^{2{x^2} + 5x - 4}}(4x + 5)$

    $f'(x) = 3(4x + 5){e^{2{x^2} + 5x - 4}}$

    Slope: $\quad {m_{tan}} = f'(0) = 3(4\cdot0 + 5){e^{2\cdot0 + 5\cdot0 - 4}}$

    $\quad{m_{tan}} = 3(5){e^{ - 4}}=15{e^{ - 4}}$

    Point: $\quad f(0)=3{e^{2\cdot0 + 5\cdot0 - 4}} = 3{e^{ - 4}}$

    $\quad(0,3{e^{ - 4}})$

    $y - 3{e^{ - 4}} = 15{e^{ - 4}}(x - 0)$

    $y - 3{e^{ - 4}} = 15{e^{ - 4}}x$

    $y = 15{e^{ - 4}}x + 3{e^{ - 4}}$

    $y = \frac{{15x}}{{{e^4}}} + \frac{3}{{{e^4}}}$

    Horizontal Tangents:

    Horizontal Tangents are found where $f'(x) = 0.$

    $3(4x + 5){e^{2{x^2} + 5x - 4}} = 0$

    Set each factor equal to zero.

    $3 = 0$

    $3 \ne 0$

    no solution

    $4x + 5 = 0$

    $4x = - 5$

    $x = - \frac{5}{4}$

    ${e^{2{x^2} + 5x - 4}} = 0$

    $\ln {e^{2{x^2} + 5x - 4}} = \ln 0$

    $ln (0)$ is undefined.

    There is only one horizontal tangent. The horizontal tangent is at $x=-\frac{5}{4}.$

  10. $f\left( x \right) = \ln \left( {1 - {x^2} + 2{x^4}} \right)$ at $x = 1$

    $f'(x) = \frac{1}{{1 - {x^2} + 2{x^4}}}\cdot( - 2x + 8{x^3})$

    $f'(x) = \frac{{ - 2x + 8{x^3}}}{{1 - {x^2} + 2{x^4}}}$

    Slope: $\quad {m_{tan}} = f'(1) = \frac{{ - 2(1) + 8{{(1)}^3}}}{{1 - {{(1)}^2} + 2{{(1)}^4}}}$

    $\quad{m_{tan}} = \frac{{ - 2 + 8}}{{1 - 1 + 2}} = \frac{6}{2} = 3$

    $\quad{m_{tan}} = 3$

    Point: $\quad f(1)=\ln(1 - {1^2} + 2{\left( 1 \right)^4}) =\ln (1 - 1 + 2)= \ln 2$

    $\quad(1,\ln 2)$

    $y - \ln 2 = 3(x - 1)$

    $y - \ln 2 = 3x - 3$

    $y = 3x - 3 + \ln 2$

    Horizontal Tangent

    Horizontal Tangents are found where $f'(x) = 0.$

    $\frac{{ - 2x + 8{x^3}}}{{1 - {x^2} + 2{x^4}}} = 0$

    $\frac{{ - 2x + 8{x^3}}}{{1 - {x^2} + 2{x^4}}} = \frac{0}{1}$

    Cross multiply.

    $- 2x + 8{x^3} = 0$

    $- 2x(1 + 4{x^2}) = 0$

    Set each factor equal to zero.

    $-2x=0$

    $x=0$

    $1 - 4{x^2} = 0$

    $x=\pm \frac{1}{2}$

    Horizontal Tangents at $x=0\;\;$ and $\;\;x=\pm \frac{1}{2}$

    Find the indicated derivative and simplify.

  11. $\frac{d}{{dt}}\;3{\left( {2{t^4}\; + {t^2}\;} \right)^{ - 5}}$

    $\frac{d}{{dt}} = - 15{\left( {2{t^4} + {t^2}} \right)^{ - 6}}(8{t^3} + 2t)$

    $ \frac{d}{{dt}} =\frac{{ - 120{t^3} - 30t}}{{{{\left( {2{t^4} + {t^2}} \right)}^6}}}$

  12. $\frac{{dh}}{{dw\;}}\quad $ if $\quad h\left( w \right) = \;\sqrt[5]{{8w - 1}}$

    $ h(w)= {\left( {8w - 1} \right)^{1/5}}$

    $\frac{{dh}}{{dw}} = \frac{1}{5}{\left( {8w - 1} \right)^{ - 4/5}}(8) $

    $ \frac{{dh}}{{dw}} =\frac{8}{{5\,\,\sqrt[5]{{{{\left( {8w - 1} \right)}^4}}}}}$

  13. $h'\left( x \right)\quad $ if $\quad h\left( x \right) = \;\frac{{{e^{4x}}}}{{{x^3}\; + 9x}}$

    $f = {e^{4x}}$

    $f' = 4{e^{4x}}$

    $g = {x^3} + 9x$

    $g' = 3{x^2} + 9$

    $h'(x) = \frac{{4{e^{4x}}({x^3} + 9x) - {e^{4x}}(3{x^2} + 9)}}{{{{\left( {{x^3} + 9x} \right)}^2}}}$

    $h'(x) = \frac{{{e^{4x}}(4{x^3} + 36x - 3{x^2} - 9)}}{{{{\left( {{x^3} + 9x} \right)}^2}}}$

    $h'(x) = \frac{{{e^{4x}}(4{x^3} - 3{x^2} + 36x - 9)}}{{{{\left( {{x^3} + 9x} \right)}^2}}}$

  14. $\frac{d}{{dx}}\;\left[ {{x^5}\;\;ln\left( {3 + {x^5}\;} \right)} \right]$

    $f = {x^5}$

    $f' = 5{x^4}$

    $g = \ln (3 + {x^5})$

    $g' =\frac{1}{{3 + {x^5}}}\cdot5{x^4} = \frac{{5{x^4}}}{{3 + {x^5}}}$

    $\frac{d}{{dx}} = 5{x^4}(\ln (3 + {x^5})) + \frac{{5{x^4}}}{{3 + {x^5}}}({x^5})$

    $\frac{d}{{dx}} = 5{x^4}\ln (3 + {x^5}) + \frac{{5{x^9}}}{{3 + {x^5}}}$

  15. $G'\left( t \right)$ if $G\left( t \right) = \;{\left( {t - {e^{9t}}} \right)^2}$

    $G'(t) = 2{\left( {t - {e^{9t}}} \right)^1}(1 - 9{e^{9t}})$

    $ G'(t) = 2(t - {e^{9t}})(1 - 9{e^{9t}})$

  16. $y'\quad $ if $\quad y = \;{\left[ {ln\left( {{x^2}\; + 7} \right)} \right]^{4/5}}$

    $y' = \frac{4}{5}{\left[ {\ln ({x^2} + 7)} \right]^{ - 1/5}}(\frac{1}{{{x^2} + 7}})(2x)$

    $y' = \frac{{8x}}{{5({x^2} + 7){{\left( {\ln ({x^2} + 7)} \right)}^{1/5}}}}$

  17. $\frac{d}{{dw}}\;\frac{1}{{{{\left( {{w^2}\; - 5} \right)}^3}}}$

    $\frac{d}{{dw}} = \left(w^2-5\right)^{-3} $

    $\frac{d}{{dw}} = - 3{\left( {{w^2} - 5} \right)^{ - 4}}(2w)$

    $\frac{d}{{dw}} = \frac{{ - 6w}}{{{{\left( {{w^2} - 5} \right)}^4}}}$

    Horizontal Tangents

    Find $f'(x)$ and simplify. Then find the equation of the tangent line to the graph of $f(x)$ at the given value of $x$. Find the values of $x$ where the tangent line is horizontal.

  18. $f\left( x \right) = {x^2}\;{\left( {3 - 2x} \right)^4}\quad\quad x = 1$

    Point: $\quad f(1) = {1^2}{\left( {3 - 2\cdot 1} \right)^4}= {1^2}{\left( 1 \right)^4} = 1$

    $(1,1)$

    $h = {x^2}$

    $h' = 2x$

    $g = {\left( {3 - 2x} \right)^4}$

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

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

    $f'(x) = {\left( {3 - 2x} \right)^3}\left[ {2x(3 - 2x) - 8{x^2}} \right]$

    $f'(x) = {\left( {3 - 2x} \right)^3}\left[ {6x - 4{x^2} - 8{x^2}} \right]$

    $f'(x) = {\left( {3 - 2x} \right)^3}( - 12{x^2} + 6x)$

    Slope: $\quad {m_{tan}} = f'(1) = {\left( {3 - 2\cdot1} \right)^3}( - 12\cdot{1^2} + 6\cdot1)$

    ${m_{tan}} = ({1^3})( - 12 + 6) = 1( - 6)=-6$

    ${m_{tan}} = - 6$

    $y - 1 = - 6(x - 1)$

    $y - 1 = - 6x + 6$

    $y = - 6x + 7$

    Horizontal Tangent

    ${\left( {3 - 2x} \right)^3}( - 12{x^2} + 6x) = 0$

    ${\left( {3 - 2x} \right)^3}(6x)( - 2x + 1) = 0$

    $3-2x = 0$

    $x = \frac{3}{2}$

    $6x = 0$

    $x=0$

    $-2x+1= 0$

    $x = \frac{1}{2}$

    Horizontal Tangents at $x=\frac{3}{2}\;\;$ and $\;x = 0\;$ and $\;x=\frac{1}{2}.$

  19. $f\left( x \right) = \;\frac{{{x^4}}}{{{{\left( {2x - 5} \right)}^2}}}\quad \quad \quad x = 2$

    Point: $\quad f(2) = \frac{{{2^4}}}{{{{\left( {2\cdot2 - 5} \right)}^2}}} = \frac{{16}}{{{{\left( { - 1} \right)}^2}}} = 16$

    $(2,16)$

    $h = {x^4}$

    $h' = 4{x^3}$

    $g ={\left( {2x - 5} \right)^2}$

    $g' =2{\left( {2x - 5} \right)^1}(2)=4(2x - 5)$

    $f'(x) = \frac{{4{x^3}{{\left( {2x - 5} \right)}^2} - {x^4}\cdot{4(2x - 5)}}}{{{{\left[ {{{\left( {2x - 5} \right)}^2}} \right]}^2}}}$

    $f'(x) = \frac{{4{x^3}{{\left( {2x - 5} \right)}^2} - 4{x^4}(2x - 5)}}{{{{\left( {2x - 5} \right)}^4}}}$

    $f'(x) = \frac{(2x-5)[{4x^3}(2x-5) - 4{x^4}]}{(2x - 5)^4}$

    $f'(x) = \frac{{4{x^3}(2x - 5) - 4{x^4}}}{{{{\left( {2x - 5} \right)}^3}}}$

    $f'(x) = \frac{{8{x^4} - 20{x^3} - 4{x^4}}}{{{{\left( {2x - 5} \right)}^3}}}$

    $f'(x) = \frac{{4{x^4} - 20{x^3}}}{{{{\left( {2x - 5} \right)}^3}}}$

    Slope: $\quad {m_{tan}} = f'(2) = \frac{{4{{(2)}^4} - 20{{(2)}^3}}}{{{{\left( {2\cdot2 - 5} \right)}^3}}}$

    ${m_{tan}} = \frac{{64 - 160}}{{{{\left( { - 1} \right)}^3}}} = \frac{{ - 96}}{{ - 1}}$

    ${m_{tan}} = 96$

    $y - 16 = 96(x - 2)$

    $y - 16 = 96x - 192$

    $y = 96x - 176$

    Horizontal Tangent

    $\frac{{4{x^4} - 20{x^3}}}{{{{\left( {2x - 5} \right)}^3}}} = \frac{0}{1}$

    $4{x^4} - 20{x^3} = 0$

    $4{x^3}(x - 5) = 0$

    Set each factor equal to zero.

    $4{x^3} = 0$

    $x = 0$

    $x - 5 = 0$

    $ x = 5$

    Horizontal Tangents at $x=0\;\;$ and $\;x = 5.$

  20. $f\left( x \right)={{e}^{\sqrt{x}}}\quad\quad$ when $x=1$

    Point: $\quad f(1) = {e^{\sqrt 1 }} = {e^1} = e$

    $(1,e)$

    $f'(x)={e^{{x^{1/2}}}}$

    $f'(x) = {e^{{x^{1/2}}}}\cdot\frac{1}{2}{x^{ - 1/2}}$

    $f'(x) = {e^{\sqrt x }}\cdot\frac{1}{{2\sqrt x }}$

    $f'(x) = \frac{{{e^{\sqrt x }}}}{{2\sqrt x }}$

    Slope: $\quad {m_{tan}} = f'(1) = \frac{{{e^{\sqrt 1 }}}}{{2\sqrt 1 }} $

    ${m_{tan}}= \frac{e}{2}$

    $y - e = \frac{e}{2}(x - 1)$

    $y - e = \frac{e}{2}x - \frac{e}{2}$

    $y = \frac{e}{2}x - \frac{e}{2} + e$

    $y = \frac{e}{2}x + \frac{e}{2}$

    Horizontal tangent

    $\frac{{{e^{\sqrt x }}}}{{2\sqrt x }} = \frac{0}{1}$

    ${e^{\sqrt x }} \ne 0$

    There are no horizontal tangents.

  21. $f\left( x \right)=~\sqrt{{{x}^{2}}~+4x+5}$ at $x = 0$

    $f(x)=(x^2+4x+5)^{ 1/2}$

    Point: $\quad f(0) = \sqrt {{0^2} + 4\cdot0 + 5} = \sqrt 5$

    $(0,\sqrt 5 )$

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

    $f'(x) = \frac{{2x + 4}}{{2\sqrt {{x^2} + 4x + 5} }}$

    $f'(x) = \frac{{2(x + 2)}}{{2\sqrt {{x^2} + 4x + 5} }} = \frac{{x + 2}}{{\sqrt {{x^2} + 4x + 5} }}$

    Slope: $\quad {m_{tan}}= f'(0) = \frac{{0 + 2}}{{\sqrt {0 + 0 + 5} }} =$

    $ {m_{tan}}=\frac{2}{{\sqrt 5 }}$

    $y - \sqrt 5 = \frac{2}{{\sqrt 5 }}(x - 0)$

    $y - \sqrt 5 = \frac{{2x}}{{\sqrt 5 }}$

    $y = \frac{{2x}}{{\sqrt 5 }} + \sqrt 5$

    Horizontal tangent

    $\frac{{x + 2}}{{\sqrt {{x^2} + 4x + 5} }} = \frac{0}{1}$

    $x + 2 = 0$

    One horizontal tangent at $\; x = - 2$

    Applications

  22. COST FUNCTION: The total cost (in hundreds of dollars) of producing x pairs of sandals per week is: $C\left( x \right) = 6 + \sqrt {3x + 25}$ when $0\; \le x\; \le 30.$

    1. Find $C'(x).$

      $C(x)=6+{{\left( 3x+25 \right)}^{{1}/{2}\;}}$

      $C'(x)=0+\frac{1}{2}{{\left( 3x+25 \right)}^{-1/2}}(3)$

      $C'(x)=\frac{3}{2\sqrt{3x+25}}$

    2. Find $C'\left( {17} \right)$ and $C'\left( {26} \right)$. Interpret the results.

      $C'(17) = \frac{3}{{2\sqrt {3\cdot17 + 25} }} = .172$

      $C'(26) = \frac{3}{{\sqrt {3\cdot26 + 25} }} = .148$

      At production level of 17 pairs per week, marginal cost is increasing by \$17.20 per pair.

      At production level of 26 pairs per week, marginal cost is increasing by \$14.80 per pair.

  23. PRICE DEMAND EQUATION: The number of large pumpkin spice drinks $(x)$ people are willing to buy per week from a local coffee shop at a price of p (in dollars) is given by:

    $$x = 1000 - 60{(p + 25)^{1/2}}$$ when $3.50\; \le p\; \le 6.25.$

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

      $x=1000-60{{\left( p+25 \right)}^{1/2}}$

      $\frac{dx}{dp}=0-60(\frac{1}{2}{{\left( p+25 \right)}^{-1/2}})(1)$

      $\frac{dx}{dp}=\frac{-30}{\sqrt{p+25}}$

    2. Find the demand and the instantaneous rate of change of demand with respect to price when the price is \$4.50. Write a brief interpretation of these results.

      $x = 1000 - 60\sqrt {4.50 + 25} = 674.12$

      $\frac{{dx}}{{dp}} = \frac{{ - 30}}{{\sqrt {4.50 + 25} }} = - 5.52$

      At \$4.50 the demand is approximately 674 drinks per week. The demand is decreasing by approximately 5 drinks per week.

  24. BIOLOGY: A yeast culture at room temperature (68°F) is placed in a refrigerator set at a constant temperature of 38°F. After t hours, the temperature, T, of the culture is given approximately by

    $T = 25{e^{ - 0.62t}} + 38$ $\quad 0\leq t\leq4.$

    What is the rate of change of temperature of the culture at the end of 1 hour? At the end of 4 hours?

    $T' = 25{e^{ - 0.62t}}( - 0.62)$

    $T' = - 15.5{e^{ - 0.62t}}$

    $T'(1) = 15.5{e^{ - 0.62(1)}} = - {8.338^ \circ }\;per\;hour$

    At the end of one hour, the temperature of the culture is decreasing at a rate of 8.338 degrees per hour.

    $T'(4) = - 15.5{e^{( - 0.62(4))}} = - {1.298^ \circ }\;per\;hour$

    At the end of five hours, the temperature of the culture is decreasing at at rate of 1.298 degrees per hour.

2.5A The Chain Rule

Practice

Find the derivative of each of the following.

  1. $p(t) = {({t^3} + 4)^5}$

    $p'(t) = 5{\left( {{t^3} + 4} \right)^4}(3{t^2})$

    $p'(t) = 15{t^2}{\left( {{t^3} + 4} \right)^4}$

  2. $f(x) = \sqrt {{x^2} - 144} $

    $ f(x)= {({x^2} - 144)^{1/2}}$

    $f'(x) = \frac{1}{2}{\left( {{x^2} - 144} \right)^{ - 1/2}}(2x)$

    $f'(x) = \frac{{2x}}{{2\sqrt {{x^2} - 144} }}$

    $f'(x) = \frac{x}{{\sqrt {{x^2} - 144} }}$

  3. $g(x) = {e^{2{x^2} - 5x + 4}}$

    $g'(x) = {e^{2{x^2} - 5x + 4}}(4x - 5)$

  4. $p(a) = \ln \;({a^4} + 4a)$

    $p'(a) = \frac{1}{{{a^4} + 4a}}(4{a^3} + 4)$

    $p'(a) = \frac{{4{a^3} + 4}}{{{a^4} + 4a}}$

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

    $y={{\left( x-{{x}^{3}} \right)}^{-1/3}}$

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

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

  6. $g(x) = 9{(2{x^2} + x - 7)^{ - 3}}$

    $g'(x)=-27{{\left( 2{{x}^{2}}+x-7 \right)}^{-4}}(4x+1)$

    $g'(x)=\frac{-27(4x+1)}{{{\left( 2{{x}^{2}}+x-7 \right)}^{4}}}$

  7. $y = {(7 - 5\ln x)^3}$

    $y'=3\left(7-5\ln x\right)^2\cdot\left(\frac{-5}x\right)$

    $y'=\frac{-15\left(7-5\ln x\right)^2}x$

  8. $q(t) = \ln (\ln \;5t)$

    $q'\left(t\right)=\frac1{\ln(5t)}\cdot\frac1{5t}\cdot5$

    $q'(t) = \frac{1}{{t\ln 5t}}$

  9. Find the equation for the tangent line to the curve $y = {\sqrt {{e^x} + 8}}$ at the point where $x = 0.$

    when $x = 0$

    $y = \sqrt {{e^0} + 8} = \sqrt {1 + 8} = \sqrt 9 = 3$

    $(0,3)$

    $y = {\left( {{e^x} + 8} \right)^{ 1/2}}$

    $y' = \frac{1}{2}{\left( {{e^x} + 8} \right)^{ - 1/2}}({e^x})$

    $y' = \frac{{{e^x}}}{{2\sqrt {{e^x} + 8} }}$

    ${m_{\tan }}= \frac{{{e^0}}}{{2\sqrt {{e^0} + 8} }} = \frac{1}{{2\sqrt 9 }} = \frac{1}{6}$

    $y - 3 = \frac{1}{6}(x - 0)$

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

    $y = \frac{1}{6}x + 3$

  10. The concentration of toxic material in a lake is related to the number of months that an manufacturing plant has been operating near the lake. This concentration of toxic material can be modeled by $A(t)={{(0.7{{t}^{1/4}}+5)}^{3}}$ where A is measured in parts per million (ppm).

    1. Find the model for the rate of change in the concentration of the toxic material in the lake.

      $A'(t)=3\left(0.7t^{1/4}+5\right)^2\left(0.7\cdot\frac14\cdot t^{-3/4}\right)$

      $A'(t)=\frac{2.1\left(0.7t^{1/4}+5\right)^2}{4t^{3/4}}$

      $A'(t)=\frac{21\left(0.7t^{1/4}+5\right)^2}{40t^{3/4}}$

    2. Find $A(20)$ and $A'(20)$and interpret the results.

      $A(20)={{(0.7{{(20)}^{1/4}}+5)}^{3}} = 272.14$

      The concentration of toxic material after 20 months is $272.14$ ppm.

      $A'(20)=\frac{21\left(0.7(20)^{1/4}+5\right)^2}{40(20)^{3/4}}$

      The concentration of toxic material after 20 months is increasing at a rate of $2.33$ ppm per month.

    3. Use the results from part b to estimate the total amount of toxic material in the lake at 21 months.

      $A(21)\approx A(20)+A'(20)$

      $A(21)\approx 272.14+2.33 = 274.47$

      The total estimated amount of toxic material in the lake at 21 months is $274.47$ ppm.

       

2.5B The Chain Rule

Practice

Find the derivative of each of the following.

  1. $p(t) = {t^2}{(5t + 1)^3}$

    $f={{t}^{2}}$

    $f'=2t$

    $g={{\left( 5t+1 \right)}^{3}}$

    $g'=3{{\left( 5t+1 \right)}^{2}}(5)$

    $g'=15{{\left( 5t+1 \right)}^{2}}$

    ​

    $p'(t) = 2t{\left( {5t + 1} \right)^3} + {t^2}(15{\left( {5t + 1} \right)^2})$

    $p'(t) = 2t{\left( {5t + 1} \right)^3} + 15{t^2}{\left( {5t + 1} \right)^2}$

    $p'(t) = {\left( {5t + 1} \right)^2}\left[ {2t(5t + 1) + 15{t^2}} \right]$

    $p'(t) = {\left( {5t + 1} \right)^2}\left[ {10{t^2} + 2t + 15{t^2}} \right]$

    $p'(t) = {\left( {5t + 1} \right)^2}(25{t^2} + 2t)$

  2. $r(x) = (2{x^2} - 3){(7x + 4)^3}$

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

    $f' = 4x$

    $g = {\left( {7x + 4} \right)^3}$

    $g' = 3{\left( {7x + 4} \right)^2}(7)$

    $g' = 21{\left( {7x + 4} \right)^2}$

    ​

    $r'(x) = 4x{\left( {7x + 4} \right)^3} + (2{x^2} - 3)\left[ {21{{\left( {7x + 4} \right)}^2}} \right]$

    $r'(x) = {\left( {7x + 4} \right)^2}\left[ {4x(7x + 4) + 21(2{x^2} - 3)} \right]$

    $r'(x) = {\left( {7x + 4} \right)^2}\left[ {28{x^2} + 16x + 42{x^2} - 63} \right]$

    $r'(x) = {\left( {7x + 4} \right)^2}\left[ {70{x^2} + 16x - 63} \right]$

  3. $p(a) = {a^3}\ln(a^5)$

    $f = {a^3}$

    $f' = 3{a^2}$

    $g = \ln ({a^5})$

    $g' = \frac{1}{{a^5}} \cdot 5{a^4}$

    $g' = \frac5a$

    ​

    $p'(a) = 3{a^2}\ln {a^5} + {a^3}(\frac5a)$

    $p'(a) = 3{a^2}\ln {a^5} + 5a^2$

    $p'(a) = a^2(3\ln {a^5} + 5)$

  4. $y =\frac{e^{x^2+x}}{4x-7}$

    $f = {e^{{x^2} + x}}$

    $f' = {e^{{x^2} + x}}(2x + 1)$

    $g = 4x-7$

    $g' =4$

    ​

    $y'=\frac{e^{x^2+x}(2x+1)(4x-7)-4e^{x^2+x}}{\left(4x-7\right)^2}$

    $y'=\frac{e^{x^2+x}\left[(2x+1)(4x-7)-4\right]}{\left(4x-7\right)^2}$

    $y'=\frac{e^{x^2+x}\left[8x^2-14x+4x-7-4\right]}{\left(4x-7\right)^2}$

    $y'=\frac{e^{x^2+x}\left[8x^2-10x-11\right]}{\left(4x-7\right)^2}$

  5. $h(x) = (6x + 5){({x^2} + 4x + 8)^{ - 2}}$

    $f = 6x + 5$

    $f' = 6$

    $g = {\left( {{x^2} + 4x + 8} \right)^{ - 2}}$

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

    ​

    $h'(x) = 6{\left( {{x^2} + 4x + 8} \right)^{ - 2}} + (6x + 5)[ - 2{\left( {{x^2} + 4x + 8} \right)^{ - 3}}(2x + 4)]$

    $h'(x) = 6{\left( {{x^2} + 4x + 8} \right)^{ - 2}} - 2(6x + 5)(2x + 4){\left( {{x^2} + 4x + 8} \right)^{ - 3}}$

    $h'(x) = \frac{6}{{{{\left( {{x^2} + 4x + 8} \right)}^2}}} - \frac{{2(6x + 5)(2x + 4)}}{{{{\left( {{x^2} + 4x + 8} \right)}^3}}}$

  6. $y =\frac{\ln\left(9x+2\right)^2}x$

    $f=\ln(9x+2)^2=2\ln\left(9x+2\right)$

    $f'=2\cdot\frac1{9x+2}\cdot9=\frac{18}{9x+2}$

    $g=x$

    $g'=1$

    ​

    $y'=\frac{\displaystyle\frac{18}{9x+2}\left(x\right)-1\cdot\ln\left(9x+2\right)^2}{x^2}$

    $y'=\frac{\displaystyle\frac{18x}{9x+2}-\ln\left(9x+2\right)^2}{x^2}$

    Multiply each term by $9x+2.$

    $y'=\frac{\displaystyle18x-\left(9x+2\right)\ln\left(9x+2\right)^2}{x^2(9x+2)}$

  7. $q(x) = \frac{e^{2x}}{e^{3x}+1}$

    $f = {e^{2x}}$

    $f' = {e^{2x}}(2)$

    $f' = 2{e^{2x}}$

    $g = {e^{3x}} + 1$

    $g' = {e^{3x}}(3)$

    $g' = 3{e^{3x}}$

    ​

    $q'(x) = \frac{2e^{2x}(e^{3x}+1)-e^{2x}(3e^{3x})}{(e^{3x}+1)^2}$

    $q'(x) = \frac{2e^{5x}+2e^{2x}-3e^{5x}}{\left(e^{3x}+1\right)^2}$

    $q'(x) = \frac{-e^{5x}+2e^{2x}}{\left(e^{3x}+1\right)^2}$

  8. $h(x) = \frac{2\sqrt x}{\left(x^2-36\right)^3}$

    $f = 2x^{1/2}$

    $f' = x^{-1/2}$

    $g = (x^2-36)^3$

    $g' = 3(x^2-36)^2 \cdot 2x=6x(x^2-36)^2$

    ​

    $h'(x) = \frac{(x^{-1/2})(x^2-36)^3-6x(x^2-36)^2(2x^{1/2})}{[(x^2-36)^3]^2}$

    $h'(x) = \frac{x^{-1/2}(x^2-36)^3-12x^{3/2}(x^2-36)^2}{(x^2-36)^6}$

    Factor $(x^2-36)^2$ from each term and reduce.

    $h'(x) = \frac{\left(x^2-36\right)^2\left[x^{-1/2}(x^2-36)-12x^{3/2}\right]}{\left(x^2-36\right)^6}$

    $h'(x) = \frac{x^{-1/2}(x^2-36)-12x^{3/2}}{(x^2-36)^4}$

    Multiply every term by $x^{1/2}$.

    $h'(x) = \frac{\left(x^2-36\right)-12x^2}{\sqrt x\left(x^2-36\right)^4}$

    $h'(x) = \frac{-11x^2-36}{\sqrt x\left(x^2-36\right)^4}$

  9. Find an equation for the tangent line to the curve $n(x) = {x^2}\ln x$ at the point where $x = e.$

    $n(e) = {e^2}\ln e= {e^2}(1) = {e^2}$

    $(e,{e^2})$

    $f = {x^2}$

    $f' = 2x$

    $g = \ln x$

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

    ​

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

    $n'(x) = 2x\ln x + x$

    ${m_{\tan }} = n'(e) = 2e\ln e + e$

    ${m_{\tan }} = 2e + e = 3e$

    ${m_{\tan }}=3e$

    $y - {e^2} = 3e(x - e)$

    $y - {e^2} = 3ex - 3{e^2}$

    $y = 3ex - 3{e^2} + {e^2}$

    $y = 3ex - 2{e^2}$

  10. The number of people in Knoxville who contract the flu can be modeled by $P(t) = \frac{{15,000}}{{50{e^{ - 0.3t}} + 1}}$, where $P$ is the number of people who contract the flu and $t$ is the number of days after the outbreak began.

    1. Find the model for the rate of change of the number of people with the flu.

      $f = 15000$

      $f' = 0$

      $g = 50{e^{ - 0.3t}} + 1$

      $g' = 50{e^{ - 0.3t}}( - 0.3) + 0$

      $g' = - 15{e^{ - 0.3t}}$

      ​

      $P'(t) = \frac{{0(50{e^{ - 0.3t}} + 1) - 15000( - 15{e^{ - 0.3t}})}}{{{{\left( {50{e^{ - 0.3t}} + 1} \right)}^2}}}$

      $P'(t) = \frac{{225000{e^{ - 0.3t}}}}{{{{\left( {50{e^{ - 0.3t}} + 1} \right)}^2}}}$

    2. Find $P(4)$ and $P'(4)$and interpret the results.

      $P(4)=934$

      $P’(4)=262.76$

      After 4 days, 934 people have the flu and 263 more will get the flu the next day.

    3. Use the results from part b to estimate the total number of people with the flu in Knoxville after 5 days.

      $P(5)\approx P(4)+P'(4)$

      $P(5)\approx934+263=1197$

      On the fifth day, approximately 1197 people will have the flu.

Unit 2 In-Class Review Problems

Find the derivative of each function. Show all of your work and simplify your answer.

  1. $h(x)=\left(\frac5{x^2}-3\right)\left(6x^2+1\right)$

    $h(x)=\left(5x^{-2}-3\right)\left(6x^2+1\right)$

    $f=5x^{-2}-3$

    $f'=-10x^{-3}$

    $g=6x^2+1$

    $g'=12x$

    ​

    $h'(x)=-10x^{-3}\left(6x^2+1\right)+12x\left(5x^{-2}-3\right) $

    $h'(x)=-60x^{-1}-10x^{-3}+60x^{-1}-36x$

    $h'(x)=\frac{-10}{x^3}-36x$

  2. $y=\sqrt[5]{\left(7x-8\right)^3}$

    $y=\left(7x-8\right)^{3/5}$

    $y'=\frac35\left(7x-8\right)^{-2/5}\left(7\right) $

    $y'=\frac{21}{5\left(7x-8\right)^{2/5}}$

    $y'=\frac{21}{5\sqrt[5]{\left(7x-8\right)^2}}$

  3. $k(n)=\frac{4n-7}{\left(2n^4+5\right)^2}$

    $f=4n-7$

    $f'=4$

    $g=\left(2n^4+5\right)^2$

    $g'=2\left(2n^4+5\right)^1\left(8n^3\right)$

    $g'=16n^3\left(2n^4+5\right)$

    ​

    $k'(n)=\frac{4\left(2n^4+5\right)^2-16n^3\left(2n^4+5\right)\left(4n-7\right)}{\left[\left(2n^4+5\right)^2\right]^2}$

    $k'(n)=\frac{4\left(2n^4+5\right)^2-16n^3\left(2n^4+5\right)\left(4n-7\right)}{\left(2n^4+5\right)^4}$

    $k'(n)=\frac{4\left(2n^4+5\right)-16n^3\left(4n-7\right)}{\left(2n^4+5\right)^3}$

    $k'(n)=\frac{8n^4+20-64n^4+112n^3}{\left(2n^4+5\right)^3}$

    $k'(n)=\frac{-56n^4+112n^3+20}{\left(2n^4+5\right)^3}$

  4. $f(x)=\ln\left(5x^2+3x\right)$

    $f'(x)=\frac1{5x^2+3x}\cdot\left(10x+3\right)$

    $f'(x)=\frac{10x+3}{5x^2+3x}$

  5. $r(w)=\left(w^2-2\right)e^{3w^2-7}$

    $f=\left(w^2-2\right)$

    $f'=2w$

    $g=e^{3w^2-7}$

    $g'=e^{3w^2-7}\left(6w\right)$

    $g'=6w\cdot e^{3w^2-7}$

    ​

    $r'\left(w\right)=2w\cdot e^{3w^2-7}+\left(w^2-2\right)\cdot6w\cdot e^{3w^2-7}$

    $r'\left(w\right)=e^{3w^2-7}\left[2w+\left(w^2-2\right)\cdot6w\right] $

    $r'\left(w\right)=e^{3w^2-7}\left[2w+6w^3-12w\right] $

    $r'\left(w\right)=e^{3w^2-7}\left[6w^3-10w\right] $

    $r'\left(w\right)=2we^{3w^2-7}\left[3w^2-5\right] $

Additional Resources