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

$\textcolor{b l u e}{f ' \left(x\right) = \frac{{x}^{2} \cdot {e}^{3 - x} + 2 x \cdot {e}^{3 - x} + 4 x}{{e}^{3 - x} + 2} ^ 2}$

#### Explanation:

Start from the given $f \left(x\right) = {x}^{2} / \left({e}^{3 - x} + 2\right)$

Use the formula $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} \left(u\right) - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{v} ^ 2$

Let $u = {x}^{2}$ and $v = {e}^{3 - x} + 2$

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{2} / \left({e}^{3 - x} + 2\right)\right)$

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

$f ' \left(x\right) = \frac{\left({e}^{3 - x} + 2\right) \cdot 2 x - {x}^{2} \left({e}^{3 - x} \left(0 - 1\right) + 0\right)}{{e}^{3 - x} + 2} ^ 2$

$f ' \left(x\right) = \frac{2 x {e}^{3 - x} + 4 x + {x}^{2} {e}^{3 - x}}{{e}^{3 - x} + 2} ^ 2$

then rearranging the numerator

$\textcolor{red}{f ' \left(x\right) = \frac{{x}^{2} {e}^{3 - x} + 2 x {e}^{3 - x} + 4 x}{{e}^{3 - x} + 2} ^ 2}$

God bless....I hope the explanation is useful.