To differentiate the function [tex]\( f(x) = \ln(x^2 - 12x + 40) \)[/tex], we will use the chain rule for differentiation.
1. Identify the outer and inner functions:
- The outer function is [tex]\( \ln(u) \)[/tex], where [tex]\( u = x^2 - 12x + 40 \)[/tex].
- The inner function is [tex]\( u = x^2 - 12x + 40 \)[/tex].
2. Differentiate the outer function with respect to the inner function:
- The derivative of [tex]\( \ln(u) \)[/tex] with respect to [tex]\( u \)[/tex] is [tex]\( \frac{1}{u} \)[/tex].
3. Differentiate the inner function with respect to [tex]\( x \)[/tex]:
- The inner function is [tex]\( u = x^2 - 12x + 40 \)[/tex].
- The derivative of [tex]\( u = x^2 - 12x + 40 \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( \frac{d}{dx}(x^2 - 12x + 40) = 2x - 12 \)[/tex].
4. Apply the chain rule:
- According to the chain rule, the derivative of [tex]\( f(x) = \ln(u) \)[/tex] with respect to [tex]\( x \)[/tex] is given by:
[tex]\[
f'(x) = \left(\frac{d}{du} \ln(u)\right) \cdot \left(\frac{du}{dx}\right)
\][/tex]
- Substitute the derivatives we found:
[tex]\[
f'(x) = \frac{1}{u} \cdot (2x - 12)
\][/tex]
- Since [tex]\( u = x^2 - 12x + 40 \)[/tex], we have:
[tex]\[
f'(x) = \frac{1}{x^2 - 12x + 40} \cdot (2x - 12)
\][/tex]
Therefore, the derivative of [tex]\( f(x) = \ln(x^2 - 12x + 40) \)[/tex] is:
[tex]\[
f'(x) = \frac{2x - 12}{x^2 - 12x + 40}
\][/tex]
