# How do you differentiate f(x)= ( -2x^2+ 3x ) / ( - e^x + 2 )  using the quotient rule?

Mar 19, 2016

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{- {e}^{x} \left(2 {x}^{2} - 7 x + 3\right) - 8 x + 6}{- {e}^{x} + 2} ^ 2$

#### Explanation:

Quotient rule states that if $f \left(x\right) = g \frac{x}{h \left(x\right)}$, then

$f ' \left(x\right) = \frac{h \left(x\right) \cdot g ' \left(x\right) - h ' \left(x\right) \cdot g \left(x\right)}{h {\left(x\right)}^{2}}$

Hence for $f \left(x\right) = \frac{- 2 {x}^{2} + 3 x}{- {e}^{x} + 2}$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\left(- {e}^{x} + 2\right) \left(- 4 x + 3\right) - \left(- {e}^{x}\right) \left(- 2 {x}^{2} + 3 x\right)}{- {e}^{x} + 2} ^ 2$

= $\frac{4 x {e}^{x} - 3 {e}^{x} - 8 x + 6 - 2 {x}^{2} {e}^{x} + 3 x {e}^{x}}{- {e}^{x} + 2} ^ 2$ or

= $\frac{- {e}^{x} \left(2 {x}^{2} - 7 x + 3\right) - 8 x + 6}{- {e}^{x} + 2} ^ 2$