Karl's Calculus Tutor - Solution to Exercise 6.2-5

Box 6.2-5: Solution to Exercise 6.2-5

The problem was to find the derivative of

   f(x)  =  b^-x²

To apply the chain rule, you need to identify what two functions f(x) is a composite of. That is we need to see f(x) as

   f(x)  =  g(h(x) )

and identify what g(x) and h(x) are. Then the chain rule tells us to find the derivative by taking

   f'(x)  =  g'(h(x) ) h'(x)

So we'll need to know g'(x) and h'(x) as well.

Do you agree that if we take

   g(x)  =  b^x

and

   h(x)  =  -x²

then the composite of these two yields f(x) = b^-x²?

Taking the derivative of g(x) using the method from this section, we have

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

And taking the derivative of h(x) using the methods we developed in the early sections on derivatives, we have

   h'(x)  =  -2x

Now simply substitute all this back into the chain rule:

   f'(x)  =  g'(h(x) ) h'(x)  =  (ln(b) b^-x²) (-2x)

That's it.