Section	3.3
                       Derivatives	of	Exponential	and
                          Logarithmic	Functions

                              V63.0121.027, Calculus	I



                                 October	22, 2009



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




                                                      .     .    .    .   .   .
Derivatives	of	Exponential	Functions

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

   Proof.
   Follow	your	nose:

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


   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:
                        (       )
                              1 n
               e = lim 1 +        = lim (1 + h)1/h
                   n→∞        n     h→0



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




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



   Question
                eh − 1
   What	is lim         ?
            h→0    h
   Answer
   If h is	small	enough, e ≈ (1 + h)1/h . So
                   [          ]h
          eh − 1    (1 + h)1/h − 1   (1 + h) − 1  h
                 ≈                 =             = =1
             h             h              h       h



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



   Question
                eh − 1
   What	is lim         ?
            h→0    h
   Answer
   If h is	small	enough, e ≈ (1 + h)1/h . So
                   [          ]h
          eh − 1    (1 + h)1/h − 1   (1 + h) − 1  h
                 ≈                 =             = =1
             h             h              h       h

                                       eh − 1
   So	in	the	limit	we	get	equality: lim       =1
                                   h→0    h
                                               .   .   .   .   .   .
Derivative	of	the	natural	exponential	function


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

   we	get:
   Theorem
                                   d x
                                      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
      x 2 ex




                            .   .   .   .   .   .
Examples

  Examples
  Find	these	derivatives:
      e3x
           2
      ex
      x 2 ex

  Solution
       d 3x
          e = 3e3x
       dx




                            .   .   .   .   .   .
Examples

  Examples
  Find	these	derivatives:
      e3x
           2
      ex
      x 2 ex

  Solution
       d 3x
          e = 3e3x
       dx
       d x2      2 d              2
          e = ex     (x2 ) = 2xex
       dx         dx




                                      .   .   .   .   .   .
Examples

  Examples
  Find	these	derivatives:
      e3x
           2
      ex
      x 2 ex

  Solution
       d 3x
          e = 3e3x
       dx
       d x2      2 d              2
          e = ex     (x2 ) = 2xex
       dx         dx
       d 2 x
          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


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



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


  Let y = ln x. Then           y
                               .
  x = ey so
       dy
         ey
          =1
       dx                                              .n x
                                                       l
       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
        x3    3x2     power	function	is	a
                      power	function	of	one
        x2    2x1
                      lower	power
        x1    1x0
        x0     0
        ?      ?
       x−1   −1x−2
       x−2   −2x−3




                       .    .   .   .   .     .
The	Tower	of	Powers


        y         y′
                        The	derivative	of	a
        x3       3x2    power	function	is	a
                        power	function	of	one
        x2       2x1
                        lower	power
        x1       1x0    Each	power	function	is
        x   0
                  0     the	derivative	of	another
                        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
        x3       3x2    power	function	is	a
                        power	function	of	one
        x2       2x1
                        lower	power
        x1       1x0    Each	power	function	is
        x   0
                  0     the	derivative	of	another
                        power	function, except
       ln x      x −1   x−1
       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
                                         d x
   Use	implicit	differentiation	to	find      a.
                                         dx




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




                                                  .   .   .   .   .   .
Other	logarithms
   Example
                                           d x
   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 =⇒    = (ln a)y = (ln a)ax
                 y dx           dx




                                                       .   .   .   .   .   .
Other	logarithms
   Example
                                           d x
   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 =⇒    = (ln a)y = (ln a)ax
                 y dx           dx

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

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

                                                        .   .   .   .      .   .
Other	logarithms

   Example
          d
   Find      loga x.
          dx




                       .   .   .   .   .   .
Other	logarithms

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




                                .   .   .   .   .   .
Other	logarithms

   Example
          d
   Find      loga x.
          dx
   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      loga x.
          dx
   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    1 1
   So      =        .
        dx   ln a x

                                                 .    .     .   .   .   .
More	examples



  Example
         d
  Find      log2 (x2 + 1)
         dx




                            .   .   .   .   .   .
More	examples



  Example
         d
  Find      log2 (x2 + 1)
         dx
  Answer

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




                                              .   .      .   .   .   .
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
   Let y =                . Find y′ .
                x−1




                                        .   .   .   .   .   .
A nasty	derivative


   Example         √
           (x2 + 1) x + 3
   Let y =                . Find 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
           √                                      √
        2x x + 3        (x2 + 1)        (x 2 + 1 ) x + 3
      =           + √                −
         (x − 1 )   2 x + 3(x − 1)           (x − 1)2




                                             .   .    .   .   .   .
Another	way


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

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


                                              .   .   .   .    .   .
Compare	and	contrast

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

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




                                            .      .   .   .     .   .
Compare	and	contrast

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

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


      Are	these	the	same?




                                            .      .   .   .     .   .
Compare	and	contrast

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

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

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




                                                .   .   .   .   .   .

Lesson 16: Derivatives of Logarithmic and Exponential Functions

  • 1.
    Section 3.3 Derivatives of Exponential and Logarithmic Functions V63.0121.027, Calculus I October 22, 2009 . . 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) a x+ h − a x f′ (x) = lim = lim h→0 h h→0 h a x a h − ax a h−1 = lim = ax · lim = ax · f′ (0). h→0 h h→0 h . . . . . .
  • 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) a x+ h − a x f′ (x) = lim = lim h→0 h h→0 h a x a h − ax a h−1 = lim = ax · lim = ax · f′ (0). h→0 h h→0 h 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: ( ) 1 n e = lim 1 + = lim (1 + h)1/h n→∞ n h→0 Question eh − 1 What is lim ? h→0 h . . . . . .
  • 7.
    The funny limit in the case of e Remember the definition of e: ( ) 1 n e = lim 1 + = lim (1 + h)1/h n→∞ n h→0 Question eh − 1 What is lim ? h→0 h Answer If h is small enough, e ≈ (1 + h)1/h . So [ ]h eh − 1 (1 + h)1/h − 1 (1 + h) − 1 h ≈ = = =1 h h h h . . . . . .
  • 8.
    The funny limit in the case of e Remember the definition of e: ( ) 1 n e = lim 1 + = lim (1 + h)1/h n→∞ n h→0 Question eh − 1 What is lim ? h→0 h Answer If h is small enough, e ≈ (1 + h)1/h . So [ ]h eh − 1 (1 + h)1/h − 1 (1 + h) − 1 h ≈ = = =1 h h h h eh − 1 So in the limit we get equality: lim =1 h→0 h . . . . . .
  • 9.
    Derivative of the natural exponential function From ( ) d x ah − 1 eh − 1 a = lim ax and lim =1 dx h→0 h h→0 h we get: Theorem d x 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 x 2 ex . . . . . .
  • 12.
    Examples Examples Find these derivatives: e3x 2 ex x 2 ex Solution d 3x e = 3e3x dx . . . . . .
  • 13.
    Examples Examples Find these derivatives: e3x 2 ex x 2 ex Solution d 3x e = 3e3x dx d x2 2 d 2 e = ex (x2 ) = 2xex dx dx . . . . . .
  • 14.
    Examples Examples Find these derivatives: e3x 2 ex x 2 ex Solution d 3x e = 3e3x dx d x2 2 d 2 e = ex (x2 ) = 2xex dx dx d 2 x 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 Lety = ln x. Then x = ey so . . . . . .
  • 17.
    Derivative of the natural logarithm function Lety = ln x. Then x = ey so dy ey =1 dx . . . . . .
  • 18.
    Derivative of the natural logarithm function Lety = ln x. Then x = ey so dy ey =1 dx dy 1 1 =⇒ = y = dx e x . . . . . .
  • 19.
    Derivative of the natural logarithm function Lety = 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 Lety = ln x. Then y . x = ey so dy ey =1 dx .n x l dy 1 1 =⇒ = y = dx e x . x . So: Fact d 1 ln x = dx x . . . . . .
  • 21.
    Derivative of the natural logarithm function Lety = ln x. Then y . x = ey so dy ey =1 dx .n x l 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 x3 3x2 power function is a power function of one x2 2x1 lower power x1 1x0 x0 0 ? ? x−1 −1x−2 x−2 −2x−3 . . . . . .
  • 23.
    The Tower of Powers y y′ The derivative of a x3 3x2 power function is a power function of one x2 2x1 lower power x1 1x0 Each power function is x 0 0 the derivative of another 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 x3 3x2 power function is a power function of one x2 2x1 lower power x1 1x0 Each power function is x 0 0 the derivative of another power function, except ln x x −1 x−1 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 d x Use implicit differentiation to find a. dx . . . . . .
  • 27.
    Other logarithms Example d x Use implicit differentiation to find a. dx Solution Let y = ax , so ln y = ln ax = x ln a . . . . . .
  • 28.
    Other logarithms Example d x 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 =⇒ = (ln a)y = (ln a)ax y dx dx . . . . . .
  • 29.
    Other logarithms Example d x 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 =⇒ = (ln a)y = (ln a)ax y dx dx Before we showed y′ = y′ (0)y, so now we know that 2h − 1 3h − 1 ln 2 = lim ≈ 0.693 ln 3 = lim ≈ 1.10 h→0 h h→0 h . . . . . .
  • 30.
    Other logarithms Example d Find loga x. dx . . . . . .
  • 31.
    Other logarithms Example d Find loga x. dx Solution Let y = loga x, so ay = x. . . . . . .
  • 32.
    Other logarithms Example d Find loga x. dx 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 loga x. dx 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 1 1 So = . dx ln a x . . . . . .
  • 34.
    More examples Example d Find log2 (x2 + 1) dx . . . . . .
  • 35.
    More examples Example d Find log2 (x2 + 1) dx Answer dy 1 1 2x = 2+1 (2x) = dx ln 2 x (ln 2)(x2 + 1) . . . . . .
  • 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 Let y = . Find y′ . x−1 . . . . . .
  • 38.
    A nasty derivative Example √ (x2 + 1) x + 3 Let y = . Find 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 √ √ 2x x + 3 (x2 + 1) (x 2 + 1 ) x + 3 = + √ − (x − 1 ) 2 x + 3(x − 1) (x − 1)2 . . . . . .
  • 39.
    Another way √ (x 2 + 1 ) x + 3 y= x−1 1 ln y = ln(x2 + 1) + ln(x + 3) − ln(x − 1) 2 1 dy 2x 1 1 = 2 + − y dx x + 1 2(x + 3) x − 1 So ( ) dy 2x 1 1 = + − y dx x2 + 1 2(x + 3) x − 1 ( ) √ 2x 1 1 (x2 + 1) x + 3 = + − x2 + 1 2(x + 3) x − 1 x−1 . . . . . .
  • 40.
    Compare and contrast Using the product, quotient, and power rules: √ √ ′ 2x x + 3 (x2 + 1) (x2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic differentiation: ( ) √ ′ 2x 1 1 (x2 + 1) x + 3 y = + − x2 + 1 2(x + 3) x − 1 x−1 . . . . . .
  • 41.
    Compare and contrast Using the product, quotient, and power rules: √ √ ′ 2x x + 3 (x2 + 1) (x2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic differentiation: ( ) √ ′ 2x 1 1 (x2 + 1) x + 3 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: √ √ ′ 2x x + 3 (x2 + 1) (x2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic differentiation: ( ) √ ′ 2x 1 1 (x2 + 1) x + 3 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: √ √ ′ 2x x + 3 (x2 + 1) (x2 + 1) x + 3 y = + √ − (x − 1) 2 x + 3(x − 1) (x − 1)2 Using logarithmic differentiation: ( ) √ ′ 2x 1 1 (x2 + 1) x + 3 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 . . . . . .