Page 89 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 89

80                                 DERIVATIVES                             [CHAP. 4



                              Since the logarithm is a continuous function, this can be written
                                                       (            u= u  )
                                                   1             u       1
                                                     log                   log e
                                                   u   a   u!0   u      ¼  u  a
                                                         lim 1 þ
                          by Problem 2.19, Chapter 2, with x ¼ u= u.
                                               d        log e du
                              Then by Problem 4.13,  ðlog uÞ¼  a  .
                                                    a
                                               dx         u  dx
                                                3
                                                     2
                     4.16. Calculate dy=dx if  (a) xy   3x ¼ xy þ 5,  (b) e xy  þ y ln x ¼ cos 2x.
                          (a)Differentiate with respect to x,considering y as a function of x.(We sometimes say that y is an implicit
                              function of x, since we cannot solve explicitly for y in terms of x.) Then
                                d  3   d   2   d      d               2 0   3            0
                               dx  ðxy Þ   dx  ð3x Þ¼  dx  ðxyÞþ dx  ð5Þ  or  ðxÞð3y y Þþðy Þð1Þ  6x ¼ðxÞðy ÞþðyÞð1Þþ 0
                                                                    2
                                                            3
                              where y ¼ dy=dx.  Solving, y ¼ð6x   y þ yÞ=ð3xy   xÞ.
                                    0
                                                    0
                               d  xy  d        d           xy        y
                                                                            0
                                                               0
                              dx      dx       dx                    x
                          ðbÞ   ðe Þþ   ðy ln xÞ¼  ðcos 2xÞ;  e ðxy þ yÞþ  þðln xÞy ¼ 2 sin 2x:
                                                            2x sin 2x þ xye xy  þ y
                              Solving;                 y ¼      2 xy
                                                        0
                                                               x e  þ x ln x
                                                                       2
                                                                   2
                                     2
                     4.17. If y ¼ coshðx   3x þ 1Þ, find  (a) dy=dx;  ðbÞ d y=dx .
                                                  2
                          (a)Let y ¼ cosh u, where u ¼ x   3x þ 1.  Then dy=du ¼ sinh u, du=dx ¼ 2x   3, and
                                            dy  dy du                         2
                                            dx  ¼  du dx  ¼ðsinh uÞð2x   3Þ¼ ð2x   3Þ sinhðx   3x þ 1Þ

                               2
                                                             2
                              d y  d    dy    d     du      d u        du   2
                                              sinh u  ¼ sinh u  þ cosh u
                          ðbÞ   2  ¼      ¼                   2
                              dx   dx dx   dx      dx       dx        dx
                                                       2        2              2     2
                                 ¼ðsinh uÞð2Þþðcosh uÞð2x   3Þ ¼ 2 sinhðx   3x þ 1Þþð2x   3Þ coshðx   3x þ 1Þ
                             2
                                  3
                     4.18. If x y þ y ¼ 2, find  (a) y ;  ðbÞ y at the point ð1; 1Þ.
                                                        00
                                                 0
                          (a)Differentiating with respect to x, x y þ 2xy þ 3y y ¼ 0 and
                                                                  2 0
                                                       2 0
                                                            2xy     1
                                                       0
                                                          x þ 3xy   2
                                                      y ¼  2    2  ¼   at ð1; 1Þ
                                                        2    2
                                  d      d    2xy                0                 0
                               00     0                ðx þ 3y Þð2xy þ 2yÞ ð2xyÞð2x þ 6yy Þ
                                  dx     dx x þ 3y                 2
                          ðbÞ  y ¼  ðy Þ¼    2   2  ¼                  2 2
                                                                 ðx þ 3y Þ
                                                                       3
                                                          1
                              Substituting x ¼ 1, y ¼ 1; and y ¼  ,we find y ¼  .
                                                                  00
                                                     0
                                                          2            8
                     MEAN VALUE THEOREMS
                     4.19. Prove Rolle’s theorem.
                          Case 1: f ðxÞ  0in ½a; bŠ.  Then f ðxÞ¼ 0for all x in ða; bÞ.
                                                     0
                          Case 2:  f ðxÞ 6  0in ½a; bŠ.Since f ðxÞ is continuous there are points at which f ðxÞ attains its maximum and
                          minimum values, denoted by M and m respectively (see Problem 3.34, Chapter 3).
                              Since f ðxÞ 6  0, at least one of the values M; m is not zero.  Suppose, for example, M 6¼ 0 and that
                          f ð Þ¼ M (see Fig. 4-9). For this case, f ð  þ hÞ @ f ð Þ.
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