# How do you use the product rule to differentiate f(x)=e^x/(3-x)?

Dec 2, 2017

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

#### Explanation:

$f \left(x\right) = {e}^{x} / \left(3 - x\right)$ Product rule: $\frac{d}{\mathrm{dx}} \left(g \cdot h \cdot\right) = g ' h + h ' g$

here $g = {e}^{x} \therefore g ' = {e}^{x} \mathmr{and} h = \frac{1}{3 - x} = {\left(3 - x\right)}^{-} 1$

$\therefore h ' = - 1 {\left(3 - x\right)}^{-} 2 \cdot \left(- 1\right) = \frac{1}{3 - x} ^ 2$

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

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

Dec 2, 2017

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

#### Explanation:

$\text{given "f(x)=g(x)h(x)" then}$

$f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$

$f \left(x\right) = {e}^{x} / \left(3 - x\right) = {e}^{x} {\left(3 - x\right)}^{-} 1$

$g \left(x\right) = {e}^{x} \Rightarrow g ' \left(x\right) = {e}^{x}$

$h \left(x\right) = {\left(3 - x\right)}^{-} 1$

$\Rightarrow h ' \left(x\right) = - {\left(3 - x\right)}^{-} 2 \left(- 1\right) = {\left(3 - x\right)}^{-} 2$

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

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = {e}^{x} {\left(3 - x\right)}^{-} 2 \left(1 + 3 - x\right)$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = \frac{{e}^{x} \left(4 - x\right)}{3 - x} ^ 2$