Karl's Calculus Tutor - Solution to Exercise 6.2-4

Box 6.2-4: Solution to Exercise 6.2-4KCT logo

© 1997 by Karl Hahn

The problem was to find the derivative of 

            x2
   f(x)  =     bx
            2
If you let 
            x2
   g(x)  =    
            2
and 
   h(x)  =  bx
then we are certainly looking for the derivative of a product, so we apply the product rule. And it says that f'(x) = g'(x)h(x) + g(x)h'(x). Using what you learned in the early sections on derivatives, we have 
   g'(x)  =  x
and using what you learned in this section, we have 
   h'(x)  =  ln(b) bx
Substituting g(x), h(x), g'(x), and h'(x) into the product rule, we have 
                                                x2
   f'(x)  =  g'(x)h(x) + g(x)h'(x)  =  x bx  +     ln(b) bx
                                                2
Observe that the right-hand expression has a common factor of bx. Factoring that out we get 
             æ           x2 ö
   f'(x)  =  ç x + ln(b)    ÷ bx
             è            2 ø
Notice that when you take the derivative of a product of an exponential with something else, the result is also a product of that same exponential with another expression. In general if 
   f(x)  =  g(x) bx
then 
  f'(x)  =  (g'(x) + ln(b)g(x)) bx
Can you see why?

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