Page 100 - Foundations Of Differential Calculus
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5. On the Differentiation of Algebraic Functions of One Variable  83
        Further examples of this kind are easily treated according to the given laws.

        162. If the quantity proposed for differentiation is the power of some func-
        tion whose differential we can find, then the preceding rules are sufficient
        to find the first differential. Let p be any function of x that is raised to
                                                                      n
        some power and whose differential is dp. Then the first differential of p is
        equal to np n−1 dp. From this we obtain the following.
                         n
          I. If y =(a + x) , then
                                              n−1
                                 dy = n (a + x)  dx.

                     2   2    2
          II. If y = a − x  , then
                                               2  2
                                 dy = −4xdx a − x   .
                     1         2  2    −1
         III. If y =      = a + x     , then
                    2
                   a + x 2
                                          −2xdx
                                   dy =         2  .
                                          2
                                               2
                                        (a + x )
                  √
                              2
         IV. If y =  a + bx + cx , then
                                       bdx +2cx dx
                                 dy = √             .
                                      2 a + bx + cx 2


                   3  4    4  2     4  4   2/3
          V. If y =  (a − x ) = a − x      , then
                                                        3
                               8  3     4   4   − 1 3  −8x dx
                         dy = − x dx a − x      = √         .
                                                       4
                                                    3
                               3                   3 a − x 4
                      1           2   − 1 2
         VI. If y = √      = 1 − x    , then
                    1 − x 2
                                          3
                                       2    −  2   xdx
                        dy = xdx 1 − x     =         √       .
                                                   2
                                              (1 − x ) 1 − x 2
                   3
                        √
        VII. If y =  a +  bx + x, then
                           √   
   √                √
                                                            √
                         dx b    (2 x)+ dx        dx b +2dx x
                   dy =                     =                     .
                                 √        2    √         √        2
                          3                       3
                         3   a +  bx + x      6 x    a +  bx + x
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