# Question #0d05a

Sep 2, 2017

$h ' ' \left(x\right) = \frac{60}{x + 5} ^ 3$

#### Explanation:

$\text{differentiate using the "color(blue)"quotient rule}$

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

$h ' \left(x\right) = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2 \leftarrow \text{ quotient rule}$

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

$g \left(x\right) = x + 5 \Rightarrow g ' \left(x\right) = 1$

$\Rightarrow h ' \left(x\right) = \frac{\left(x + 5\right) \left(2 x - 1\right) - \left({x}^{2} - x\right)}{x + 5} ^ 2$

$\textcolor{w h i t e}{\Rightarrow h ' \left(x\right)} = \frac{{x}^{2} + 10 x - 5}{x + 5} ^ 2$

$\text{to obtain "h''(x)" differentiate } h ' \left(x\right)$
$\text{using the "color(blue)"quotient rule}$

$f \left(x\right) = {x}^{2} + 10 x - 5 \Rightarrow f ' \left(x\right) = 2 x + 10 = 2 \left(x + 5\right)$

$g \left(x\right) = {\left(x + 5\right)}^{2} \Rightarrow g ' \left(x\right) = 2 \left(x + 5\right)$

$\Rightarrow h ' ' \left(x\right) = \frac{2 {\left(x + 5\right)}^{3} - 2 \left(x + 5\right) \left({x}^{2} + 10 x - 5\right)}{x + 5} ^ 4$

$\textcolor{w h i t e}{\Rightarrow h ' ' \left(x\right)} = \frac{\left(x + 5\right) \left[2 {\left(x + 5\right)}^{2} - 2 \left({x}^{2} + 10 x - 5\right)\right)}{x + 5} ^ 4$

$\textcolor{w h i t e}{\Rightarrow h ' ' \left(x\right)} = \frac{60}{x + 5} ^ 3 \to \left(A\right)$