Page 101 - Foundations Of Differential Calculus
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84    5. On the Differentiation of Algebraic Functions of One Variable
                        1
       VIII. If y =   √        , then since
                         2
                   x +  a − x 2
                                              −xdx
                                     2   2
                                d. a − x = √         ,
                                               2
                                              a − x 2
             we have
                                
   √
                                                          √
                                     2
                    −dx +(xdx)      a − x 2       xdx − dx a − x 2
                                                             2
               dy =          √         2           √         2 √      ,
                                2   2       =         2   2     2    2
                          x +  a − x            x +  a − x     a − x
             or
                                           √         3
                                              2    2
                                    dx x −   a − x
                               dy =          2  √     .
                                                 2
                                           2
                                       2
                                    (2x − a )   a − x 2

                                          3


                   4      1    3      2  2
         IX. If y =   1 − √ +    (1 − x )  ,welet
                           x
                            1                	        2
                                             3
                                                    2
                           √ = p      and      (1 − x ) = q;
                            x

                      4          3
             since y =  (1 − p + q) ,wehave
                                        −3dp +3dq
                                   dy = √         .
                                         4
                                        4 1 − p + q
             From previous work we have
                               −dx                  −4xdx
                         dp =   √      and    dq = √        .
                                                     3
                              2x x                 3 1 − x 2
             When these results are substituted, we have
                                       √            
√
                                   
                 3     2
                              (3dx) (2x x) − (4xdx)    1 − x
                         dy =                               .

                                    4    1    3        2
                                                     2
                                  4  1 − √ +    (1 − x )
                                          x
        In a similar way, by substituting individual letters for terms to be com-
        posed, we can easily find the differentials of this kind of function.
        163. If the quantity that is to be differentiated is the product of two or
        more functions of x whose differentials are known, the most convenient
        method for finding the differential is as follows. Let p and q be functions of
        x with differentials dp and dq already known. When we substitute x+dx for
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