Solution to Exercise 8.5-2KCT logo

© 1999 by Karl Hahn

The problem was to use logarithmic differentiation or the shortcut method to find the derivative of

            x2e-xsin(2x)
   f(x)  =              
               1 + x2
Again we have the product of four functions:
   g1(x)  =  x2          g1'(x)  =  2x

   g2(x)  =  e-x         g2'(x)  =  -e-x

   g3(x)  =  sin(2x)     g3'(x)  =  2 cos(2x)

                1                      -2x
   g4(x)  =              g4'(x)  =           
             1 + x2                 (1 + x2)2

Step 1: Take the natural log of both sides.

                 æ  x2e-xsin(2x)  ö
   ln(f(x))  = lnç                ÷
                 è     1 + x2     ø

Step 2: Apply the log identity.

   ln(f(x))  =  ln(x2)  +  ln(e-x)  +  ln(sin(2x))  -  ln(1 + x2)
which you can simplify further to
   ln(f(x))  =  2 ln(x)  -  x  +  ln(sin(2x))  -  ln(1 + x2)

Step 3: Use the chain rule to take the derivative of both sides.

   f'(x)     2           2 cos(2x)       2x
          =     -  1  +             -        
    f(x)     x            sin(2x)      1 + x2

Step 4: Multiply through by f(x).

             æ 2           2 cos(2x)       2x   ö
   f'(x)  =  ç    -  1  +             -         ÷ f(x)
             è x            sin(2x)      1 + x2 ø

Step 5: Substitute the original product for f(x).

             æ 2           2 cos(2x)       2x   ö x2e-xsin(2x)
   f'(x)  =  ç    -  1  +             -         ÷             
             è x            sin(2x)      1 + x2 ø    1 + x2

Step 6 (optional): Multiply it through.

             2xe-xsin(2x)     x2e-xsin(2x)
   f'(x)  =                -                +
                1 + x2           1 + x2


             2x2e-xcos(2x)     2x3e-xsin(2x)
                            -               
                 1 + x2          (1 + x2)2
If you used the shortcut method you would have arrived at this point in one step.


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