Box 6.2-4: Solution to Exercise 6.2-4 © 1997 by Karl Hahn

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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 b^x. Factoring that out we get

f'(x) =

x + ln(b)

x2 2

bx

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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