TheQuotientRule
                                                        mc-TY-quotient-2009-1
        Aspecial rule, the quotient rule, exists for differentiating quotients of two functions. This unit
        illustrates this rule.
        In order to master the techniques explained here it is vital that you undertake plenty of practice
        exercises so that they become second nature.
        After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
          • state the quotient rule
          • differentiate quotients of functions
                                   Contents
        1. Introduction                                           2
                                              du   dv
        2. The quotient rule: if y = u then dy = vdx −udx         2
                                 v      dx      v2
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          1. Introduction
          Functions often come as quotients, by which we mean one function divided by another function.
          For example,
                                                  y = cosx
                                                         2
                                                       x
                              u                                      2
          Wewrite this as y =   where we identify u as cosx and v as x .
                              v
          There is a formula we can use to differentiate a quotient - it is called the quotient rule. In this
          unit we will state and use the quotient rule.
          2. The quotient rule
          The rule states:
                                                      KeyPoint
           The quotient rule: if y = u then
                                     v
                                                       du    dv
                                                dy = vdx −udx
                                                dx       v2
          Let’s see how the formula works when we try to differentiate y = cosx.
                                                                          2
                                                                         x
          Example
          Suppose we want to differentiate y = cosx.
                                                2
                                               x
                                                2
          Wehave identified u as cosx and v as x . So
                                                               2
                                             u=cosx       v = x
          Wenowwrite down the derivatives of these two functions.
                                          du =−sinx        dv =2x
                                          dx               dx
          Wenowput all these results into the given formula:
                                                      du    dv
                                               dy = vdx −udx
                                               dx       v2
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         Quote the formula everytime so that you get to know it.
                                           2
                                   dy = x ·(−sinx)−cosx·2x
                                                    2 2
                                   dx             (x )
         Notice that there is a minus sign and an x in both terms of the numerator (the top line). So we
         can take out a common factor of −x.
                                     dy = −x(xsinx+2cosx)
                                                    4
                                    dx             x
                                        = −(xsinx+2cosx)
                                                    3
                                                   x
         by cancelling the factor of x in the numerator and the denominator. We have found the required
         derivative.
         Example
                                          2
         Suppose we want to differentiate y = x +6.
                                         2x−7
                                                   2
         Werecognise this as a quotient and identify u as x +6 and v as 2x−7.
                                          2
                                      u=x +6      v = 2x−7
         Differentiating
                                        du = 2x    dv =2
                                        dx         dx
         Quoting the formula:
                                                du   dv
                                          dy = vdx −udx
                                          dx      v2
         So
                                                        2
                                  dy = (2x−7)·2x−(x +6)·2
                                                       2
                                  dx            (2x−7)
                                             2        2
                                      = 2(2x −7x−x −6)
                                                     2
                                              (2x−7)
                                            2
                                      = 2(x −7x−6)
                                                  2
                                           (2x−7)
         In the following Example we will use the quotient rule to establish another result.
         Example
         Suppose we want to differentiate y = tanx.
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           Recall that tanx = sinx so we have a quotient in which
                              cosx
                                              u=sinx        v = cosx
           So
                                            du =cosx       dv = −sinx
                                            dx             dx
           Quoting the formula:
                                                         du    dv
                                                  dy = vdx −udx
                                                 dx        v2
           So
                                       dy = cosx·cosx−sinx·(−sinx)
                                                              2
                                       dx                  cos x
                                                   2       2
                                            = cos x+sin x
                                                    cos2x
           The top line can be simplified using the standard result that cos2x + sin2x = 1. So
                                                  dy =       1
                                                              2
                                                  dx       cos x
           This can be written as sec2x because the function secx is defined to be   1 .
                                                                                  cosx
           Example
           Suppose we want to differentiate y = secx.
           The function secx is defined to be   1 , that is, a quotient.
                                             cosx
           Taking
                                                u = 1     v = cosx
                                             du =0       dv = −sinx
                                             dx          dx
           Quoting the formula:
                                                         du    dv
                                                  dy = vdx −udx
                                                 dx        v2
           So
                                           dy = cosx·0−1·(−sinx)
                                                              2
                                          dx               cos x
                                               = sinx
                                                   cos2x
           Wecan write this answer in an alternative form:
                                               dy =       1  · sinx
                                               dx       cosx cosx
                                                    = secxtanx
           Wenowhave another standard result: if y = secx then dy = secxtanx.
                                                                  dx
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