differentiate. y = \\ln \\left(7x^2 - 9x + 1\ ight) y = \\square

$differentiate. y = \\ln \\left(7x^2 - 9x + 1\ ight) y = \\square$

differentiate.
y = \ln \left(7x^2 - 9x + 1\
ight)

y = \square

Let $ u = 7x^2 - 9x + 1 $, so $ y = \ln(u) $. The derivative of $ \ln(u) $ is $ \frac{1}{u} \cdot u' $.

$ u' = \frac{d}{dx}(7x^2 - 9x + 1) = 14x - 9 $.

$ y' = \frac{1}{7x^2 - 9x + 1} \cdot (14x - 9) $.

$ \frac{14x - 9}{7x^2 - 9x + 1} $