Solution to Exercise 6.2-2 © 1997 by Karl Hahn

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The first part of the problem had you apply the chain rule to find the derivative of

   f(x)  =  bcx

Recall that the chain rule says that if  f(x) = g(h(x) ),  then  f'(x) = g'(h(x) )h'(x).  So let  g(h) = b^h  and let  h(x) = cx.  Then according to what you've learned about the derivatives of exponentials,

   g'(h)  =  ln(b) bh

and from our earliest discussion of derivatives you have

   h'(x)  =  c

Applying the chain rule to the composite you get

   f'(x)  =  g'(h(x) )h'(x)  =  ln(b) bcx × c  =  c ln(b) bcx

The second part of the problem asks you to take the derivative of

   (bc)x

To do that, we simply observe that this is just another exponential, but it uses (b^c) as its base. Applying what we know about taking the derivative of exponentials we have

   f'(x)  =  ln(bc) (bc)x

But we know that

   (bc)x  =  bcx

Substituting we get

   f'(x)  =  ln(bc) bcx

Since we took the derivative of two expressions that are equal, the derivatives must be equal as well. So that gives us

   c ln(b) bcx  =  ln(bc) bcx

Dividing out the b^cx from both sides we have

   c ln(b)  =  ln(bc)

which is another familiar property of logarithms.

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