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Sullivan  04 apcalc4e 45342 ch02 166 233 5pp  August 7, 2023  12:54



               206     Chapter 2 • The Derivative and Its Properties

                                                    2 Differentiate the Quotient of Two Functions

                                                   The derivative of the quotient of two functions is not equal to the quotient of their
                                                   derivatives. Instead, the derivative of the quotient of two functions is found using the
                                                   Quotient Rule.


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                                                    THEOREM Quotient Rule
                                                                                                   f (x)
                                                    If two functions f and g are differentiable and if F(x) =  , g(x) 6= 0, then F
                                                                                                   g(x)
                                                    is differentiable, and the derivative of the quotient F is
                                                                                  ′   ′             ′
                                                                            f (x)    f (x)g(x) − f (x)g (x)
                                                                     ′
                                                                    F (x) =       =             2
                                                                            g(x)           [g(x)]
                IN WORDS The derivative of a quotient of
                 two functions is the derivative of the
                                                    In Leibniz notation, the Quotient Rule has the form
                 numerator times the denominator, minus the
                 numerator times the derivative of the

                 denominator, all divided by the denominator                        d  f (x) g(x) − f (x)  d  g(x)
                 squared. That is,                            d        d     f (x)     dx             dx
                                                                F(x) =          =
                             ′                                                                    2
                                ′
                           f   f g − f g  ′                   dx       dx g(x)               [g(x)]
                             =
                          g       g 2
                                                   Proof We use the definition of a derivative (Form 2) to find F (x).
                                                                                                     ′
                                                                                              f (x + h)  f (x)
                                                                                                      −
                                                                          F(x + h) − F(x)     g(x + h)  g(x)
                                                               F (x) = lim              = lim
                                                                ′
                                                                      h→0       h       ↑  h→0      h
                                                                                           f (x)
                                                                                     F(x) =
                                                                                           g(x)
                                                                          f (x + h)g(x) − f (x)g(x + h)
                                                                    = lim
                                                                      h→0      h[g(x + h)g(x)]
                                                      We write F in an equivalent form that contains the difference quotients for f and g
                                                                ′
                                                   by subtracting and adding f (x)g(x) to the numerator.
                                                                     f (x + h)g(x) − f (x)g(x) + f (x)g(x) − f (x)g(x + h)
                                                          F (x) = lim
                                                            ′
                                                                 h→0                h[g(x + h)g(x)]
                                                   Now group and factor the numerator.
                                                               [ f (x + h) − f (x)]g(x) − f (x)[g(x + h) − g(x)]
                                                     F (x) = lim
                                                      ′
                                                            h→0             h[g(x + h)g(x)]

                                                                 f (x + h) − f (x)         g(x + h) − g(x)
                                                                                g(x) − f (x)
                                                                       h                         h
                                                          = lim
                                                            h→0                g(x + h)g(x)

                                                                 f (x + h) − f (x)                      g(x + h) − g(x)
                                                            lim                 · lim g(x) − lim f (x) · lim
                                                            h→0        h         h→0       h→0     h→0        h
                                                          =
                                                                                lim g(x + h) · lim g(x)
                                                                                h→0        h→0
                RECALL Since g is differentiable, it is     f (x)g(x) − f (x)g (x)
                                                             ′
                                                                            ′
                 continuous; so, lim g(x + h) = g(x).     =            2
                           h→0                                    [g(x)]




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