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Solutions to Worksheet for Sections 3.1–3.3
   Derivatives of Exponential and Logarithmic Functions
                                   V63.0121, Calculus I
                                          Summer 2010

Find the derivatives of the following functions.
             2
                 −3x
1. y = e2x

   Solution. We have
                           dy      2     d               2
                              = e2x −3x    2x2 − 3x = e2x −3x (4x − 3)
                           dx           dx


2. y = 62x

   Solution. We have
                                   y = (ln 6) · 62x · 2 = (2 ln 6)62x



3. y = ln(x3 + 9)

   Solution. We have
                                 dy     1   d          3x2
                                    = 3       x3 + 9 = 3
                                 dx  x + 9 dx         x +9


4. y = log3 ex

   Solution. By brute force we have
                                                  1              1
                                      y =                · ex =
                                             (ln 3) · ex        ln 3
   But slightly more elegantly, we notice that
                                                      ln ex    x
                                          log3 ex =         =
                                                       ln 3   ln 3
        dy    1
   So      =      makes a lot of sense.
        dx   ln 3

                                                 1
5. y =    log10 3θ2 −θ

   Solution.
                  dy   1            2      −1/2         1                     2
                     =     log10 3θ −θ          ·         θ 2 −θ
                                                                 · (ln 3) · 3θ −θ (2θ − 1)
                  dx   2                          (ln 10)3
                         ln 3         2θ − 1
                     =         ·
                       2 ln 10    log10 3θ2 −θ

   There’s some simplifications we could do before differentiation, however.
                                                    2
                                                        −θ
                                            ln 3θ                 θ2 − θ ln 3    ln 3 2
                               θ 2 −θ
                         log10 3        =                     =               =       θ −θ
                                               ln 10                 ln 10      ln 10
   So

                                                              ln 3
                                                y=                 · θ2 − θ
                                                             ln 10
                                                              ln 3   2θ − 1
                                               y =                 · √
                                                             ln 10 2 θ2 − θ




6. y = sin2 x + 2sin x

   Solution.
                                        y = 2 sin x cos x + 2sin x · ln 2 · cos x



Use logarithmic differentiation to find the derivatives of the following functions.

7. y = x x2 − 1

   Solution. We have
                                                     1
                                                       ln(x2 − 1)
                                            ln y = ln x +
                                                     2
                                         1 dy   1 1      2x
                                              = + · 2
                                         y dx   x 2 x −1
                                           dy              1      x
                                              = x x2 − 1     + 2
                                           dx              x x −1



8. y =    (x − 1)(x − 2)(x − 3)

                                                             2
Solution. We have
                              1
                       ln y =   (ln(x − 1) + ln(x − 2) + ln(x − 3))
                              2
                       1 dy   1     1        1        1
                            =           +        +
                       y dx   2 x−1 x−2 x−3
                         dy   1                             1       1   1
                            =     (x − 1)(x − 2)(x − 3)         +     +
                         dx   2                           x−1 x−2 x−3



         x(x − 1)3/2
9. y =     √
             x+1

  Solution. We have
                                       3            1
                            ln y = ln x +ln(x − 1) − ln(x + 1)
                                       2            2
                           1 dy  1 3       1     1    1
                                = + ·          − ·
                           y dx  x 2 x−1 2 x+1
                             dy  x(x − 1)3/2 1 3        1    1 1
                                = √              + ·       − ·
                             dx      x+1       x 2 x−1 2 x+1




                                             3

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Lesson 8: Derivatives of Logarithmic and Exponential Functions (worksheet solutions)

  • 1. Solutions to Worksheet for Sections 3.1–3.3 Derivatives of Exponential and Logarithmic Functions V63.0121, Calculus I Summer 2010 Find the derivatives of the following functions. 2 −3x 1. y = e2x Solution. We have dy 2 d 2 = e2x −3x 2x2 − 3x = e2x −3x (4x − 3) dx dx 2. y = 62x Solution. We have y = (ln 6) · 62x · 2 = (2 ln 6)62x 3. y = ln(x3 + 9) Solution. We have dy 1 d 3x2 = 3 x3 + 9 = 3 dx x + 9 dx x +9 4. y = log3 ex Solution. By brute force we have 1 1 y = · ex = (ln 3) · ex ln 3 But slightly more elegantly, we notice that ln ex x log3 ex = = ln 3 ln 3 dy 1 So = makes a lot of sense. dx ln 3 1
  • 2. 5. y = log10 3θ2 −θ Solution. dy 1 2 −1/2 1 2 = log10 3θ −θ · θ 2 −θ · (ln 3) · 3θ −θ (2θ − 1) dx 2 (ln 10)3 ln 3 2θ − 1 = · 2 ln 10 log10 3θ2 −θ There’s some simplifications we could do before differentiation, however. 2 −θ ln 3θ θ2 − θ ln 3 ln 3 2 θ 2 −θ log10 3 = = = θ −θ ln 10 ln 10 ln 10 So ln 3 y= · θ2 − θ ln 10 ln 3 2θ − 1 y = · √ ln 10 2 θ2 − θ 6. y = sin2 x + 2sin x Solution. y = 2 sin x cos x + 2sin x · ln 2 · cos x Use logarithmic differentiation to find the derivatives of the following functions. 7. y = x x2 − 1 Solution. We have 1 ln(x2 − 1) ln y = ln x + 2 1 dy 1 1 2x = + · 2 y dx x 2 x −1 dy 1 x = x x2 − 1 + 2 dx x x −1 8. y = (x − 1)(x − 2)(x − 3) 2
  • 3. Solution. We have 1 ln y = (ln(x − 1) + ln(x − 2) + ln(x − 3)) 2 1 dy 1 1 1 1 = + + y dx 2 x−1 x−2 x−3 dy 1 1 1 1 = (x − 1)(x − 2)(x − 3) + + dx 2 x−1 x−2 x−3 x(x − 1)3/2 9. y = √ x+1 Solution. We have 3 1 ln y = ln x +ln(x − 1) − ln(x + 1) 2 2 1 dy 1 3 1 1 1 = + · − · y dx x 2 x−1 2 x+1 dy x(x − 1)3/2 1 3 1 1 1 = √ + · − · dx x+1 x 2 x−1 2 x+1 3