Page 118 - Foundations Of Differential Calculus
P. 118

6. On the Differentiation of Transcendental Functions  101
        182. By means of this rule the differential of the logarithm of any given
        function of x whatsoever is easily found, which will be clear from the fol-
        lowing examples.
          I. If y =ln x, then
                                            dx
                                       dy =   .
                                            x
                                n
                      n
          II. If y =ln x ,welet x = p, so that y =ln p and dy = dp/p. But
             dp = nx n−1 dx, so that
                                           ndx
                                      dy =     .
                                            x
             The same result can be found from the nature of logarithms; since
                n
             ln x = n ln x,wehave
                                                 ndx
                                     n
                                d ln x = nd ln x =   .
                                                  x
                          2
         III. If y =ln 1+ x , then
                                           2xdx
                                     dy =       .
                                          1+ x 2
                        1
         IV. If y =ln √     , since
                      1 − x 2
                                               1        2
                                         2
                           y = − ln  1 − x = − ln 1 − x   ,
                                               2
             we see that
                                           xdx
                                     dy =       .
                                          1 − x 2
                        x                   1       2
          V. If y =ln √     , since y =ln x −  2  ln 1+ x ,wehave
                      1+ x 2
                                  dx    xdx        dx
                             dy =    −       =          .
                                                      2
                                  x    1+ x 2  x (1 + x )
                         √
                               2
         VI. If y =ln x +  1+ x ,wehave
                                 
√                    √
                         dx + xdx   1+ x 2     xdx + dx 1+ x 2
                    dy =        √         =      √        √      ;
                            x +  1+ x 2       x +  1+ x 2  1+ x 2
             but since both numerator and denominator of this fraction are divis-
                       √
                             2
             ible by x +  1+ x ,wehave
                                            dx
                                    dy = √       .
                                           1+ x 2
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