Another ln Problem

[Jason76]

\(\displaystyle f(x) = \ln(x^{2} - 3x + 1)\)

\(\displaystyle f'(x) = \ln[(u) 2x - 3]\)

\(\displaystyle f'(x) = \ln[(x^{2} - 3x + 1) 2x - 3]\)

 

[lookagain]

\(\displaystyle f(x) = \ln(x^{2} - 3x + 1)\)

\(\displaystyle f'(x) = \ln[(u) 2x - 3]\)

\(\displaystyle f'(x) = \ln[(x^{2} - 3x + 1) 2x - 3]\)

No, it equals \(\displaystyle \ \dfrac{2x - 3}{x^2 - 3x + 1}.\)