# What is the first and second derivative of 1/(x^2-x+2)?

Jul 2, 2015

$f ' \left(x\right) = \frac{1 - 2 x}{{\left({x}^{2} - x + 2\right)}^{2}}$ and $f ' ' \left(x\right) = \frac{6 {x}^{2} - 6 x - 2}{{\left({x}^{2} - x + 2\right)}^{3}}$

#### Explanation:

Use the Quotient Rule $\frac{d}{\mathrm{dx}} \left(g \frac{x}{h \left(x\right)}\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{{\left(h \left(x\right)\right)}^{2}}$ to get

$f ' \left(x\right) = \frac{0 \cdot \left({x}^{2} - x + 2\right) - 1 \cdot \left(2 x - 1\right)}{{\left({x}^{2} - x + 2\right)}^{2}} = \frac{1 - 2 x}{{\left({x}^{2} - x + 2\right)}^{2}}$

Use the Quotient Rule again, along with the Chain Rule d/dx(g(h(x))=g'(h(x))h'(x) to get

$f ' ' \left(x\right) = \frac{{\left({x}^{2} - x + 2\right)}^{2} \cdot \left(- 2\right) - \left(1 - 2 x\right) \cdot 2 \left({x}^{2} - x + 2\right) \cdot \left(2 x - 1\right)}{{\left({x}^{2} - x + 2\right)}^{4}}$

Now simplify

$f ' ' \left(x\right) = \frac{\left({x}^{2} - x + 2\right) \left(- 2 {x}^{2} + 2 x - 4 + 2 {\left(2 x - 1\right)}^{2}\right)}{{\left({x}^{2} - x + 2\right)}^{4}}$

$= \frac{- 2 {x}^{2} + 2 x - 4 + 8 {x}^{2} - 8 x + 2}{{\left({x}^{2} - x + 2\right)}^{3}} = \frac{6 {x}^{2} - 6 x - 2}{{\left({x}^{2} - x + 2\right)}^{3}}$