Karl's Calculus Tutor - Solution to Exercise 6.2-2

Box 6.2-2: Solution to Exercise 6.2-2

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

   f(x)  =  b^cx

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

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

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) b^cx * c  =  c ln(b) b^cx

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

   (b^c)^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(b^c) (b^c)^x

But we know that

   (b^c)^x  =  b^cx

Substituting we get

   f'(x)  =  ln(b^c) b^cx

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) b^cx  =  ln(b^c) b^cx

Dividing out the b^cx from both sides we have

   c ln(b)  =  ln(b^c)

which is another familiar property of logrithms.

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