Page 130 - Foundations Of Differential Calculus
P. 130

6. On the Differentiation of Transcendental Functions  113
        199. Although the arc whose sine or cosine or tangent or cotangent or
        secant or cosecant or finally versed sine is given is a transcendental quantity,
        nevertheless its differential when divided by dx is an algebraic quantity. It
        follows that its second differential, its third, fourth, and so forth, when
        divided by the appropriate power of dx, are also algebraic. In order that
        this differentiation might better be seen, we adjoin some examples.
                          √                    √
                                 2
                                                     2
          I. If y = arcsin 2x 1 − x ,welet p =2x 1 − x , so that y = arcsin p
                               2

             and dy = dp/ 1 − p . But then
                                            2

                                     2
                                          2x dx    2dx 1 − 2x 2
                       dp =2dx 1 − x − √         =   √
                                          1 − x 2      1 − x 2
                       2
                                2
             and   1 − p =1 − 2x , so when these values are substituted we have
                                           2dx
                                    dy = √       .
                                           1 − x 2
                                     √
                                           2
             From this it is clear that 2x 1 − x is the sine of twice the arc where
             x is the sine of the original arc. Hence if y = 2 arcsin x, then dy =
                 √
                       2
             2dx/ 1 − x .
          II. If
                                            1 − x 2
                                   y = arcsin     ,
                                            1+ x 2
             we let
                                          1 − x 2
                                      p =      ,
                                          1+ x 2
             so that
                            −4xdx                          2x
                                                     2
                      dp =        2    and       1 − p =     2  .
                                 2
                           (1 + x )                      1+ x
             Since
                                            dp
                                                 ,
                                    dy =        2
                                           1 − p
             we have
                                          −2dx
                                     dy =       .
                                          1+ x 2

         III. If y = arcsin  (1 − x)/2, we let p =  (1 − x)/2, so that

                             2     1+ x                  −dx
                         1 − p =            and    dp = 	     .
                                     2                     1−x
                                                        4
                                                            2
             It follows that
                                      dp        −dx
                                            = √       .
                               dy =       2          2
                                      1 − p   2 1 − x
   125   126   127   128   129   130   131   132   133   134   135