How do I find the derivative of #f(x)=(16x^2+1) ln (x-3)#?

1 Answer
Dec 22, 2015

This question will demand the use of chain rule and product rule, as well as the knowledge of how to derivate a logarithmic function.

Explanation:

First of all, we have a product of two terms:

  • #16x^2+1#
  • #ln(x-3)#

However, one cannot derivate straightforward a logarithmic function with a function in it. Instead, we must apply chain rule and rename #u=x-3# so now we can proceed. The chain rule states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#, so:

#(dy)/(dx)=(1/u)(1)=1/u=1/(x-3)#

Fine. Now, we can apply the product rule, which states that, for #f(x)=g(x)*h(x)#, #f'(x)=g'(x)*h(x)+g(x)*h'(x)#, as follows:

(where the symbol #'# indicates the first derivative)

#f'(x)=32xln(x-3)+(16x^2+1)(1/(x-3))#

#f'(x)=32xln(x-3)+(16x^2+1)/(x-3)#

The struggle to develop and try to simplify it is not worth it. This is your final answer!