Solution to Exercise 8.5-2


© 1999 by Karl Hahn 		KCT logo

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)


   g4(x)  =


     1
        
  1 + x2


      g4'(x)  =


    -2x
           
  (1 + x2)2

Step 1: Take the natural log of both sides. 

   ln(f(x))  =  ln



x2e-xsin(2x)
            
   1 + x2

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)
          =
    f(x)

  2
     -  1  +
  x

  2 cos(2x)
             -
   sin(2x)

    2x
        
  1 + x2

Step 4: Multiply through by  f(x).  

   f'(x)  =

  

2           2 cos(2x)
   -  1  +             -
x            sin(2x)

    2x
        
  1 + x2



 f(x)

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

   f'(x)  =

  

2           2 cos(2x)
   -  1  +             -
x            sin(2x)

    2x
        
  1 + x2



 x2e-xsin(2x)
             
    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.

Return to Main Text

You can email me by clicking this button:

  Use your own emailer     Use form