# How do you find the second derivative of ln(x^2-3x+3) ?

Nov 1, 2016

Firstly, use implicit differentiation...

$y = \ln \left({x}^{2} - 3 x + 3\right)$

${e}^{y} = {x}^{2} - 3 x + 3$

${e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x - 3}{e} ^ y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x - 3}{{x}^{2} - 3 x + 3}$

Now use the quotient rule...

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{\left({x}^{2} - 3 x + 3\right) \cdot 2 - \left(2 x - 3\right) \left(2 x - 3\right)}{{\left({x}^{2} - 3 x + 3\right)}^{2}}$

And simplify the fraction...

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 {x}^{2} - 6 x + 6 - \left(4 {x}^{2} - 12 x + 9\right)}{{\left({x}^{2} - 3 x + 3\right)}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 {x}^{2} - 6 x + 6 - 4 {x}^{2} + 12 x - 9}{{\left({x}^{2} - 3 x + 3\right)}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 2 {x}^{2} + 6 x - 3}{{\left({x}^{2} - 3 x + 3\right)}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{6 x - 2 {x}^{2} - 3}{{\left({x}^{2} - 3 x + 3\right)}^{2}}$