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Section 3.3
                 Derivatives of Exponential and
                     Logarithmic Functions

                                    V63.0121, Calculus I


                                     March 10/11, 2009


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                  Quiz 3 this week: Covers Sections 2.1–2.4
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Image credit: heipei
                                                           .   .      .   .   .   .
Outline


   Derivative of the natural exponential function
      Exponential Growth

   Derivative of the natural logarithm function

   Derivatives of other exponentials and logarithms
      Other exponentials
      Other logarithms

   Logarithmic Differentiation
      The power rule for irrational powers



                                                    .   .   .   .   .   .
Derivatives of Exponential Functions


   Fact
   If f(x) = ax , then f′ (x) = f′ (0)ax .




                                             .   .   .   .   .   .
Derivatives of Exponential Functions


   Fact
   If f(x) = ax , then f′ (x) = f′ (0)ax .

   Proof.
   Follow your nose:

                          f(x + h) − f(x)         ax+h − ax
             f′ (x) = lim                 = lim
                                 h                    h
                      h→0                     h→0
                          ax ah − ax            ah − 1
                                                        = ax · f′ (0).
                                     = ax · lim
                    = lim
                               h                   h
                      h→0                   h→0




                                                       .    .    .       .   .   .
Derivatives of Exponential Functions


   Fact
   If f(x) = ax , then f′ (x) = f′ (0)ax .

   Proof.
   Follow your nose:

                          f(x + h) − f(x)         ax+h − ax
             f′ (x) = lim                 = lim
                                 h                    h
                      h→0                     h→0
                          ax ah − ax            ah − 1
                                                        = ax · f′ (0).
                                     = ax · lim
                    = lim
                               h                   h
                      h→0                   h→0



   To reiterate: the derivative of an exponential function is a constant
   times that function. Much different from polynomials!


                                                       .    .    .       .   .   .
The funny limit in the case of e
   Remember the definition of e:
                         (        )
                                1n
                                    = lim (1 + h)1/h
                e = lim 1 +
                                n
                    n→∞               h→0



   Question
              eh − 1
   What is lim       ?
                 h
          h→0




                                             .    .    .   .   .   .
The funny limit in the case of e
   Remember the definition of e:
                         (        )
                                1n
                                    = lim (1 + h)1/h
                e = lim 1 +
                                n
                    n→∞               h→0



   Question
                eh − 1
   What is lim         ?
                   h
            h→0

   Answer
   If h is small enough, e ≈ (1 + h)1/h . So
                    [          ]h
                     (1 + h)1/h − 1
           eh − 1                     (1 + h) − 1  h
                  ≈                 =             = =1
              h             h              h       h



                                               .   .   .   .   .   .
The funny limit in the case of e
   Remember the definition of e:
                         (        )
                                1n
                                    = lim (1 + h)1/h
                e = lim 1 +
                                n
                    n→∞               h→0



   Question
                eh − 1
   What is lim         ?
                   h
            h→0

   Answer
   If h is small enough, e ≈ (1 + h)1/h . So
                    [          ]h
                     (1 + h)1/h − 1
           eh − 1                     (1 + h) − 1  h
                  ≈                 =             = =1
              h             h              h       h
                                        eh − 1
                                               =1
   So in the limit we get equality: lim
                                           h
                                    h→0
                                                    .   .   .   .   .   .
Derivative of the natural exponential function



   From
                     (                )
                             ah − 1                      eh − 1
             dx
                                          ax
                a=                                              =1
                         lim                   and   lim
             dx                 h                           h
                         h→0                         h→0

   we get:
   Theorem
                                  dx
                                     e = ex
                                  dx




                                                       .   .   .     .   .   .
Exponential Growth


      Commonly misused term to say something grows exponentially
      It means the rate of change (derivative) is proportional to the
      current value
      Examples: Natural population growth, compounded interest,
      social networks




                                                .    .    .    .    .   .
Examples

  Examples
  Find these derivatives:
       e3x
            2
       ex
       x2 ex




                            .   .   .   .   .   .
Examples

  Examples
  Find these derivatives:
       e3x
            2
       ex
       x2 ex

  Solution
       d 3x
          e = 3ex
       dx




                            .   .   .   .   .   .
Examples

  Examples
  Find these derivatives:
       e3x
            2
       ex
       x2 ex

  Solution
       d 3x
          e = 3ex
       dx
       d x2     2d            2
          e = ex (x2 ) = 2xex
       dx        dx




                                  .   .   .   .   .   .
Examples

  Examples
  Find these derivatives:
       e3x
            2
       ex
       x2 ex

  Solution
       d 3x
          e = 3ex
       dx
       d x2     2d            2
          e = ex (x2 ) = 2xex
       dx        dx
       d 2x
          x e = 2xex + x2 ex
       dx


                                  .   .   .   .   .   .
Outline


   Derivative of the natural exponential function
      Exponential Growth

   Derivative of the natural logarithm function

   Derivatives of other exponentials and logarithms
      Other exponentials
      Other logarithms

   Logarithmic Differentiation
      The power rule for irrational powers



                                                    .   .   .   .   .   .
Derivative of the natural logarithm function


  Let y = ln x. Then
  x = ey so




                                       .   .   .   .   .   .
Derivative of the natural logarithm function


  Let y = ln x. Then
  x = ey so
             dy
        ey      =1
             dx




                                       .   .   .   .   .   .
Derivative of the natural logarithm function


  Let y = ln x. Then
  x = ey so
        dy
        ey =1
        dx
        dy   1  1
     =⇒    = y=
        dx  e   x




                                       .   .   .   .   .   .
Derivative of the natural logarithm function


  Let y = ln x. Then
  x = ey so
        dy
         ey=1
        dx
        dy   1  1
     =⇒    = y=
        dx  e   x
  So:
  Fact
         d         1
            ln x =
         dx        x



                                       .   .   .   .   .   .
Derivative of the natural logarithm function

                                 y
                                 .
  Let y = ln x. Then
  x = ey so
        dy
         ey=1
        dx                                             l
                                                       .n x
        dy   1  1
     =⇒    = y=
        dx  e   x
                                 .                     x
                                                       .
  So:
  Fact
         d         1
            ln x =
         dx        x



                                       .   .   .   .   .      .
Derivative of the natural logarithm function

                                 y
                                 .
  Let y = ln x. Then
  x = ey so
        dy
         ey=1
        dx                                             l
                                                       .n x
        dy   1  1                                       1
     =⇒    = y=                                        .
        dx  e   x                                       x
                                 .                     x
                                                       .
  So:
  Fact
         d         1
            ln x =
         dx        x



                                       .   .   .   .   .      .
The Tower of Powers


               y′
         y
                      The derivative of a power
        x3     3x2    function is a power
                      function of one lower
        x2     2x1
                      power
        x1     1x0
        x0      0
         ?      ?
        x−1   −1x−2
        x−2   −2x−3




                        .   .    .   .    .       .
The Tower of Powers


                 y′
         y
                        The derivative of a power
        x3       3x2    function is a power
                        function of one lower
        x2       2x1
                        power
        x1       1x0    Each power function is
                        the derivative of another
            0
        x         0
                        power function, except
                 x−1    x−1
         ?
        x−1     −1x−2
        x−2     −2x−3




                          .    .   .    .    .      .
The Tower of Powers


                 y′
         y
                        The derivative of a power
        x3       3x2    function is a power
                        function of one lower
        x2       2x1
                        power
        x1       1x0    Each power function is
                        the derivative of another
            0
        x         0
                        power function, except
                 x−1    x−1
        ln x
        x−1     −1x−2   ln x fills in this gap
                        precisely.
        x−2     −2x−3




                          .     .    .     .    .   .
Outline


   Derivative of the natural exponential function
      Exponential Growth

   Derivative of the natural logarithm function

   Derivatives of other exponentials and logarithms
      Other exponentials
      Other logarithms

   Logarithmic Differentiation
      The power rule for irrational powers



                                                    .   .   .   .   .   .
Other logarithms

   Example
                                         dx
   Use implicit differentiation to find      a.
                                         dx




                                                 .   .   .   .   .   .
Other logarithms

   Example
                                         dx
   Use implicit differentiation to find      a.
                                         dx
   Solution
   Let y = ax , so
                             ln y = ln ax = x ln a




                                                     .   .   .   .   .   .
Other logarithms

   Example
                                         dx
   Use implicit differentiation to find      a.
                                         dx
   Solution
   Let y = ax , so
                               ln y = ln ax = x ln a
   Differentiate implicitly:

                     1 dy           dy
                                       = (ln a)y = (ln a)ax
                          = ln a =⇒
                     y dx           dx




                                                       .   .   .   .   .   .
Other logarithms

   Example
                                         dx
   Use implicit differentiation to find      a.
                                         dx
   Solution
   Let y = ax , so
                               ln y = ln ax = x ln a
   Differentiate implicitly:

                     1 dy           dy
                                       = (ln a)y = (ln a)ax
                          = ln a =⇒
                     y dx           dx

   Before we showed y′ = y′ (0)y, so now we know that

                      2h − 1                               3h − 1
                             ≈ 0.693                              ≈ 1.10
          ln 2 = lim                         ln 3 = lim
                         h                                    h
                  h→0                                  h→0


                                                        .    .   .   .     .   .
Other logarithms

   Example
          d
   Find     log x.
          dx a




                     .   .   .   .   .   .
Other logarithms

   Example
          d
   Find     log x.
          dx a
   Solution
   Let y = loga x, so ay = x.




                                .   .   .   .   .   .
Other logarithms

   Example
          d
   Find     log x.
          dx a
   Solution
   Let y = loga x, so ay = x. Now differentiate implicitly:

                             dy        dy    1         1
                  (ln a)ay      = 1 =⇒    =y      =
                             dx        dx  a ln a   x ln a




                                                       .      .   .   .   .   .
Other logarithms

   Example
          d
   Find     log x.
          dx a
   Solution
   Let y = loga x, so ay = x. Now differentiate implicitly:

                              dy        dy    1         1
                   (ln a)ay      = 1 =⇒    =y      =
                              dx        dx  a ln a   x ln a
   Another way to see this is to take the natural logarithm:

                                                              ln x
                     ay = x =⇒ y ln a = ln x =⇒ y =
                                                              ln a
        dy    11
           =
   So               .
        dx   ln a x

                                                       .       .     .   .   .   .
More examples



  Example
         d
           log (x2 + 1)
  Find
         dx 2




                          .   .   .   .   .   .
More examples



  Example
         d
           log (x2 + 1)
  Find
         dx 2
  Answer

                  dy    1      1               2x
                     =            (2x) =
                              2+1        (ln 2)(x2 + 1)
                  dx   ln 2 x




                                                .   .     .   .   .   .
Outline


   Derivative of the natural exponential function
      Exponential Growth

   Derivative of the natural logarithm function

   Derivatives of other exponentials and logarithms
      Other exponentials
      Other logarithms

   Logarithmic Differentiation
      The power rule for irrational powers



                                                    .   .   .   .   .   .
A nasty derivative


   Example         √
           (x2 + 1) x + 3
                          . Find y′ .
   Let y =
                x−1




                                        .   .   .   .   .   .
A nasty derivative


   Example         √
           (x2 + 1) x + 3
                          . Find y′ .
   Let y =
                x−1
   Solution
   We use the quotient rule, and the product rule in the numerator:
               [√                                      ]           √
       (x − 1) 2x x + 3 + (x2 + 1) 1 (x + 3)−1/2 − (x2 + 1) x + 3(1)
   ′                                    2
   y=
                                       (x − 1)2
          √                                          √
                          (x2 + 1)         (x2 + 1) x + 3
       2x x + 3
                  +√                    −
     =
        (x − 1)                                (x − 1)2
                     2 x + 3(x − 1)




                                              .   .   .    .   .       .
Another way


                              √
                      (x2 + 1) x + 3
                  y=
                            x−1
                                  1
               ln y = ln(x + 1) + ln(x + 3) − ln(x − 1)
                          2
                                  2
              1 dy       2x       1        1
                                       −
                    =2       +
                      x + 1 2(x + 3) x − 1
              y dx

  So
               (                           )
        dy           2x       1      1
                                  −
           =             +                     y
                   x2 + 1 2(x + 3) x − 1
        dx
                                                       √
               (                           )
                                               (x2 + 1) x + 3
                     2x       1      1
                                  −
          =              +
                   x2 + 1 2(x + 3) x − 1            x−1


                                                   .   .   .   .   .   .
Compare and contrast

      Using the product, quotient, and power rules:
                  √                                     √
                                 (x2 + 1)       (x2 + 1) x + 3
                2x x + 3
            ′
                          +√                 −
           y=
                 (x − 1)                            (x − 1)2
                             2 x + 3(x − 1)

      Using logarithmic differentiation:
                                                   √
                (                          )2
                                            (x + 1) x + 3
                     2x          1       1
             ′
                                       −
           y=             +
                   x2 + 1 2(x + 3) x − 1        x−1




                                             .    .   .    .     .   .
Compare and contrast

      Using the product, quotient, and power rules:
                  √                                     √
                                 (x2 + 1)       (x2 + 1) x + 3
                2x x + 3
            ′
                          +√                 −
           y=
                 (x − 1)                            (x − 1)2
                             2 x + 3(x − 1)

      Using logarithmic differentiation:
                                                   √
                (                          )2
                                            (x + 1) x + 3
                     2x          1       1
             ′
                                       −
           y=             +
                   x2 + 1 2(x + 3) x − 1        x−1


      Are these the same?




                                             .    .   .    .     .   .
Compare and contrast

      Using the product, quotient, and power rules:
                  √                                     √
                                 (x2 + 1)       (x2 + 1) x + 3
                2x x + 3
            ′
                          +√                 −
           y=
                 (x − 1)                            (x − 1)2
                             2 x + 3(x − 1)

      Using logarithmic differentiation:
                                                   √
                (                          )2
                                            (x + 1) x + 3
                     2x          1       1
             ′
                                       −
           y=             +
                   x2 + 1 2(x + 3) x − 1        x−1


      Are these the same?
      Which do you like better?




                                             .    .   .    .     .   .
Compare and contrast

      Using the product, quotient, and power rules:
                  √                                     √
                                 (x2 + 1)       (x2 + 1) x + 3
                2x x + 3
            ′
                          +√                 −
           y=
                 (x − 1)                            (x − 1)2
                             2 x + 3(x − 1)

      Using logarithmic differentiation:
                                                   √
                (                          )2
                                            (x + 1) x + 3
                     2x          1       1
             ′
                                       −
           y=             +
                   x2 + 1 2(x + 3) x − 1        x−1


      Are these the same?
      Which do you like better?
      What kinds of expressions are well-suited for logarithmic
      differentiation?

                                               .    .    .   .    .   .
Derivatives of powers




   Let y = xx . Which of these is true?
   (A) Since y is a power function, y′ = x · xx−1 = xx .
   (B) Since y is an exponential function, y′ = (ln x) · xx
   (C) Neither




                                                    .      .   .   .   .   .
Derivatives of powers




   Let y = xx . Which of these is true?
   (A) Since y is a power function, y′ = x · xx−1 = xx .
   (B) Since y is an exponential function, y′ = (ln x) · xx
   (C) Neither




                                                    .      .   .   .   .   .
It’s neither! Or both?



   If y = xx , then

                       ln y = x ln x
                      1 dy        1
                            = x · + ln x = 1 + ln x
                      y dx        x
                        dy
                            = xx + (ln x)xx
                        dx
   Each of these terms is one of the wrong answers!




                                                .     .   .   .   .   .
Derivative of arbitrary powers


   Fact (The power rule)
   Let y = xr . Then y′ = rxr−1 .




                                    .   .   .   .   .   .
Derivative of arbitrary powers


   Fact (The power rule)
   Let y = xr . Then y′ = rxr−1 .

   Proof.

                            y = xr =⇒ ln y = r ln x
   Now differentiate:
                                1 dy     r
                                      =
                                 y dx   x
                                   dy     y
                                      = r = rxr−1
                              =⇒
                                   dx     x




                                                      .   .   .   .   .   .

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Lesson 16: Derivatives of Exponential and Logarithmic Functions

  • 1. Section 3.3 Derivatives of Exponential and Logarithmic Functions V63.0121, Calculus I March 10/11, 2009 Announcements Quiz 3 this week: Covers Sections 2.1–2.4 Get half of all unearned ALEKS points by March 22 . . Image credit: heipei . . . . . .
  • 2. Outline Derivative of the natural exponential function Exponential Growth Derivative of the natural logarithm function Derivatives of other exponentials and logarithms Other exponentials Other logarithms Logarithmic Differentiation The power rule for irrational powers . . . . . .
  • 3. Derivatives of Exponential Functions Fact If f(x) = ax , then f′ (x) = f′ (0)ax . . . . . . .
  • 4. Derivatives of Exponential Functions Fact If f(x) = ax , then f′ (x) = f′ (0)ax . Proof. Follow your nose: f(x + h) − f(x) ax+h − ax f′ (x) = lim = lim h h h→0 h→0 ax ah − ax ah − 1 = ax · f′ (0). = ax · lim = lim h h h→0 h→0 . . . . . .
  • 5. Derivatives of Exponential Functions Fact If f(x) = ax , then f′ (x) = f′ (0)ax . Proof. Follow your nose: f(x + h) − f(x) ax+h − ax f′ (x) = lim = lim h h h→0 h→0 ax ah − ax ah − 1 = ax · f′ (0). = ax · lim = lim h h h→0 h→0 To reiterate: the derivative of an exponential function is a constant times that function. Much different from polynomials! . . . . . .
  • 6. The funny limit in the case of e Remember the definition of e: ( ) 1n = lim (1 + h)1/h e = lim 1 + n n→∞ h→0 Question eh − 1 What is lim ? h h→0 . . . . . .
  • 7. The funny limit in the case of e Remember the definition of e: ( ) 1n = lim (1 + h)1/h e = lim 1 + n n→∞ h→0 Question eh − 1 What is lim ? h h→0 Answer If h is small enough, e ≈ (1 + h)1/h . So [ ]h (1 + h)1/h − 1 eh − 1 (1 + h) − 1 h ≈ = = =1 h h h h . . . . . .
  • 8. The funny limit in the case of e Remember the definition of e: ( ) 1n = lim (1 + h)1/h e = lim 1 + n n→∞ h→0 Question eh − 1 What is lim ? h h→0 Answer If h is small enough, e ≈ (1 + h)1/h . So [ ]h (1 + h)1/h − 1 eh − 1 (1 + h) − 1 h ≈ = = =1 h h h h eh − 1 =1 So in the limit we get equality: lim h h→0 . . . . . .
  • 9. Derivative of the natural exponential function From ( ) ah − 1 eh − 1 dx ax a= =1 lim and lim dx h h h→0 h→0 we get: Theorem dx e = ex dx . . . . . .
  • 10. Exponential Growth Commonly misused term to say something grows exponentially It means the rate of change (derivative) is proportional to the current value Examples: Natural population growth, compounded interest, social networks . . . . . .
  • 11. Examples Examples Find these derivatives: e3x 2 ex x2 ex . . . . . .
  • 12. Examples Examples Find these derivatives: e3x 2 ex x2 ex Solution d 3x e = 3ex dx . . . . . .
  • 13. Examples Examples Find these derivatives: e3x 2 ex x2 ex Solution d 3x e = 3ex dx d x2 2d 2 e = ex (x2 ) = 2xex dx dx . . . . . .
  • 14. Examples Examples Find these derivatives: e3x 2 ex x2 ex Solution d 3x e = 3ex dx d x2 2d 2 e = ex (x2 ) = 2xex dx dx d 2x x e = 2xex + x2 ex dx . . . . . .
  • 15. Outline Derivative of the natural exponential function Exponential Growth Derivative of the natural logarithm function Derivatives of other exponentials and logarithms Other exponentials Other logarithms Logarithmic Differentiation The power rule for irrational powers . . . . . .
  • 16. Derivative of the natural logarithm function Let y = ln x. Then x = ey so . . . . . .
  • 17. Derivative of the natural logarithm function Let y = ln x. Then x = ey so dy ey =1 dx . . . . . .
  • 18. Derivative of the natural logarithm function Let y = ln x. Then x = ey so dy ey =1 dx dy 1 1 =⇒ = y= dx e x . . . . . .
  • 19. Derivative of the natural logarithm function Let y = ln x. Then x = ey so dy ey=1 dx dy 1 1 =⇒ = y= dx e x So: Fact d 1 ln x = dx x . . . . . .
  • 20. Derivative of the natural logarithm function y . Let y = ln x. Then x = ey so dy ey=1 dx l .n x dy 1 1 =⇒ = y= dx e x . x . So: Fact d 1 ln x = dx x . . . . . .
  • 21. Derivative of the natural logarithm function y . Let y = ln x. Then x = ey so dy ey=1 dx l .n x dy 1 1 1 =⇒ = y= . dx e x x . x . So: Fact d 1 ln x = dx x . . . . . .
  • 22. The Tower of Powers y′ y The derivative of a power x3 3x2 function is a power function of one lower x2 2x1 power x1 1x0 x0 0 ? ? x−1 −1x−2 x−2 −2x−3 . . . . . .
  • 23. The Tower of Powers y′ y The derivative of a power x3 3x2 function is a power function of one lower x2 2x1 power x1 1x0 Each power function is the derivative of another 0 x 0 power function, except x−1 x−1 ? x−1 −1x−2 x−2 −2x−3 . . . . . .
  • 24. The Tower of Powers y′ y The derivative of a power x3 3x2 function is a power function of one lower x2 2x1 power x1 1x0 Each power function is the derivative of another 0 x 0 power function, except x−1 x−1 ln x x−1 −1x−2 ln x fills in this gap precisely. x−2 −2x−3 . . . . . .
  • 25. Outline Derivative of the natural exponential function Exponential Growth Derivative of the natural logarithm function Derivatives of other exponentials and logarithms Other exponentials Other logarithms Logarithmic Differentiation The power rule for irrational powers . . . . . .
  • 26. Other logarithms Example dx Use implicit differentiation to find a. dx . . . . . .
  • 27. Other logarithms Example dx Use implicit differentiation to find a. dx Solution Let y = ax , so ln y = ln ax = x ln a . . . . . .
  • 28. Other logarithms Example dx Use implicit differentiation to find a. dx Solution Let y = ax , so ln y = ln ax = x ln a Differentiate implicitly: 1 dy dy = (ln a)y = (ln a)ax = ln a =⇒ y dx dx . . . . . .
  • 29. Other logarithms Example dx Use implicit differentiation to find a. dx Solution Let y = ax , so ln y = ln ax = x ln a Differentiate implicitly: 1 dy dy = (ln a)y = (ln a)ax = ln a =⇒ y dx dx Before we showed y′ = y′ (0)y, so now we know that 2h − 1 3h − 1 ≈ 0.693 ≈ 1.10 ln 2 = lim ln 3 = lim h h h→0 h→0 . . . . . .
  • 30. Other logarithms Example d Find log x. dx a . . . . . .
  • 31. Other logarithms Example d Find log x. dx a Solution Let y = loga x, so ay = x. . . . . . .
  • 32. Other logarithms Example d Find log x. dx a Solution Let y = loga x, so ay = x. Now differentiate implicitly: dy dy 1 1 (ln a)ay = 1 =⇒ =y = dx dx a ln a x ln a . . . . . .
  • 33. Other logarithms Example d Find log x. dx a Solution Let y = loga x, so ay = x. Now differentiate implicitly: dy dy 1 1 (ln a)ay = 1 =⇒ =y = dx dx a ln a x ln a Another way to see this is to take the natural logarithm: ln x ay = x =⇒ y ln a = ln x =⇒ y = ln a dy 11 = So . dx ln a x . . . . . .
  • 34. More examples Example d log (x2 + 1) Find dx 2 . . . . . .
  • 35. More examples Example d log (x2 + 1) Find dx 2 Answer dy 1 1 2x = (2x) = 2+1 (ln 2)(x2 + 1) dx ln 2 x . . . . . .
  • 36. Outline Derivative of the natural exponential function Exponential Growth Derivative of the natural logarithm function Derivatives of other exponentials and logarithms Other exponentials Other logarithms Logarithmic Differentiation The power rule for irrational powers . . . . . .
  • 37. A nasty derivative Example √ (x2 + 1) x + 3 . Find y′ . Let y = x−1 . . . . . .
  • 38. A nasty derivative Example √ (x2 + 1) x + 3 . Find y′ . Let y = x−1 Solution We use the quotient rule, and the product rule in the numerator: [√ ] √ (x − 1) 2x x + 3 + (x2 + 1) 1 (x + 3)−1/2 − (x2 + 1) x + 3(1) ′ 2 y= (x − 1)2 √ √ (x2 + 1) (x2 + 1) x + 3 2x x + 3 +√ − = (x − 1) (x − 1)2 2 x + 3(x − 1) . . . . . .
  • 39. Another way √ (x2 + 1) x + 3 y= x−1 1 ln y = ln(x + 1) + ln(x + 3) − ln(x − 1) 2 2 1 dy 2x 1 1 − =2 + x + 1 2(x + 3) x − 1 y dx So ( ) dy 2x 1 1 − = + y x2 + 1 2(x + 3) x − 1 dx √ ( ) (x2 + 1) x + 3 2x 1 1 − = + x2 + 1 2(x + 3) x − 1 x−1 . . . . . .
  • 40. Compare and contrast Using the product, quotient, and power rules: √ √ (x2 + 1) (x2 + 1) x + 3 2x x + 3 ′ +√ − y= (x − 1) (x − 1)2 2 x + 3(x − 1) Using logarithmic differentiation: √ ( )2 (x + 1) x + 3 2x 1 1 ′ − y= + x2 + 1 2(x + 3) x − 1 x−1 . . . . . .
  • 41. Compare and contrast Using the product, quotient, and power rules: √ √ (x2 + 1) (x2 + 1) x + 3 2x x + 3 ′ +√ − y= (x − 1) (x − 1)2 2 x + 3(x − 1) Using logarithmic differentiation: √ ( )2 (x + 1) x + 3 2x 1 1 ′ − y= + x2 + 1 2(x + 3) x − 1 x−1 Are these the same? . . . . . .
  • 42. Compare and contrast Using the product, quotient, and power rules: √ √ (x2 + 1) (x2 + 1) x + 3 2x x + 3 ′ +√ − y= (x − 1) (x − 1)2 2 x + 3(x − 1) Using logarithmic differentiation: √ ( )2 (x + 1) x + 3 2x 1 1 ′ − y= + x2 + 1 2(x + 3) x − 1 x−1 Are these the same? Which do you like better? . . . . . .
  • 43. Compare and contrast Using the product, quotient, and power rules: √ √ (x2 + 1) (x2 + 1) x + 3 2x x + 3 ′ +√ − y= (x − 1) (x − 1)2 2 x + 3(x − 1) Using logarithmic differentiation: √ ( )2 (x + 1) x + 3 2x 1 1 ′ − y= + x2 + 1 2(x + 3) x − 1 x−1 Are these the same? Which do you like better? What kinds of expressions are well-suited for logarithmic differentiation? . . . . . .
  • 44. Derivatives of powers Let y = xx . Which of these is true? (A) Since y is a power function, y′ = x · xx−1 = xx . (B) Since y is an exponential function, y′ = (ln x) · xx (C) Neither . . . . . .
  • 45. Derivatives of powers Let y = xx . Which of these is true? (A) Since y is a power function, y′ = x · xx−1 = xx . (B) Since y is an exponential function, y′ = (ln x) · xx (C) Neither . . . . . .
  • 46. It’s neither! Or both? If y = xx , then ln y = x ln x 1 dy 1 = x · + ln x = 1 + ln x y dx x dy = xx + (ln x)xx dx Each of these terms is one of the wrong answers! . . . . . .
  • 47. Derivative of arbitrary powers Fact (The power rule) Let y = xr . Then y′ = rxr−1 . . . . . . .
  • 48. Derivative of arbitrary powers Fact (The power rule) Let y = xr . Then y′ = rxr−1 . Proof. y = xr =⇒ ln y = r ln x Now differentiate: 1 dy r = y dx x dy y = r = rxr−1 =⇒ dx x . . . . . .