Karl's Calculus Tutor - Solution to Exercise 6.2-1

Box 6.2-1: Solution to Exercise 6.2-1

The first part of the problem was to find the derivative of

   f(x)  =  a^x b^x

using the product rule. This expression is clearly a product.

Step 1: What are the two factors in the product? If, as we have done in the past, we want to label the two factors, g(x) and h(x), then we have

   g(x)  =  a^x

and

   h(x)  =  b^x

Step 2: What are the derivatives of the two factors? Using what we learned about taking the derivative of an exponential in section 6.2, we also have

   g'(x)  =  ln(a) a^x

and

   h'(x)  =  ln(b) b^x

Step 3: Apply the product rule. The product rule tells us that if f(x) = g(x)h(x), then f'(x) = g'(x)h(x) + g(x)h'(x). Substituting the expression we have for g(x), h(x), g'(x), and h'(x), we get

   f'(x)  =  g'(x)h(x) + g(x)h'(x)  =  ln(a) a^x b^x  +  ln(b) a^x b^x

Step 4: Simplify. Observe that the expression on the right has a common factor of a^x b^x. If you factor that out you get

   f'(x)  =  (ln(a) + ln(b) ) a^x b^x

Now on to the second part of the problem. Recall that at the end of section 6.1, you proved that

   a^x b^x  =  (ab)^x

So another expression for f(x) is (ab)^x. The second part of the problem asks you to take the derivative of this expression. Applying what you learned about taking the derivative of exponentials, we get

   f'(x)  =  ln(ab) (ab)^x

Since both a^x b^x and (ab)^x are expressions for the identical function, the derivatives that we find for them must also be equal. Hence

   ln(ab) (ab)^x  =  (ln(a) + ln(b) ) a^x b^x

and if you substitute (ab)^x for a^x b^x (which you can do because you already established that they are equal), you get

   ln(ab) (ab)^x  =  (ln(a) + ln(b) ) (ab)^x

Now simply divide out the (ab)^x from both sides and you get

   ln(ab)  =  ln(a) + ln(b)

which is a well known property of logrithms.

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