# How do you find the derivative of h(p) = (1+p^3) / (5p + (3p^2)?

Jun 4, 2015

Here, we can use the product rule, which states that, be $y = f \left(x\right) g \left(x\right)$, then:

$y ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

So, rewriting our quotient, we have $\left(1 + {p}^{3}\right) {\left(5 p + 3 {p}^{2}\right)}^{-} 1$.

To derivate the denominator, we'll use the chain rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$. Here, we'll rename $u = 5 p + 3 {p}^{2}$.

Let's go!

$\frac{\mathrm{dh} \left(p\right)}{\mathrm{dp}} = \left(3 {p}^{2}\right) {\left(5 p + 3 {p}^{2}\right)}^{-} 1 + \left(1 + {p}^{3}\right) \left(- 1\right) \left(5 + 6 p\right) {\left(5 p + 3 {p}^{2}\right)}^{-} 2$

$\frac{\mathrm{dh} \left(p\right)}{\mathrm{dp}} = \frac{3 {p}^{2}}{5 p + 3 {p}^{2}} - \frac{5 + 5 {p}^{3} + 6 p + 6 {p}^{4}}{5 p + 3 {p}^{2}} ^ 2$

l.c.d: ${\left(5 p + 3 {p}^{2}\right)}^{2}$

$\frac{\mathrm{dh} \left(p\right)}{\mathrm{dp}} = \frac{\left(3 {p}^{2}\right) \left(5 p + 3 {p}^{2}\right) - \left(5 + 5 {p}^{3} + 6 p + 6 {p}^{4}\right)}{5 p + 3 {p}^{2}} ^ 2$

$\frac{\mathrm{dh} \left(p\right)}{\mathrm{dp}} = \frac{15 {p}^{3} + 9 {p}^{4} - 5 - 5 {p}^{3} - 6 p - 6 {p}^{4}}{5 p + 3 {p}^{2}} ^ 2$

$\frac{\mathrm{dh} \left(p\right)}{\mathrm{dp}} = \frac{3 {p}^{4} + 10 {p}^{3} - 6 p - 5}{25 {p}^{2} + 30 {p}^{3} + 9 {p}^{4}}$

$\frac{\mathrm{dh} \left(p\right)}{\mathrm{dp}} = \frac{3 {p}^{4} + 10 {p}^{3} - 6 p - 5}{{p}^{2} \left(25 + 30 p + 9 {p}^{2}\right)}$

We can factor the denominator's parenthesis by finding its roots:

$\frac{- 30 \pm \sqrt{900 - 4 \left(9\right) \left(25\right)}}{18}$
${p}_{1} = {p}_{2} = - \frac{5}{3}$, thus $\left(3 p + 5\right) \left(3 p + 5\right) = \left(25 + 30 p + 9 {p}^{2}\right)$

$\frac{\mathrm{dh} \left(p\right)}{\mathrm{dp}} = \textcolor{g r e e n}{\frac{3 {p}^{4} + 10 {p}^{3} - 6 p - 5}{{p}^{2} {\left(3 p + 5\right)}^{2}}}$