# How do you differentiate z=t^2/((t-4)(2-t^3)) using the quotient rule?

Mar 3, 2018

$= \frac{2 {t}^{5} - 4 {t}^{4} + 2 {t}^{2} - 16 t}{{\left(2 - {t}^{3}\right)}^{2} \cdot {\left(t - 4\right)}^{2}}$

#### Explanation:

Quotient Rule states that:

$\frac{d}{\mathrm{dx}} \left(\frac{\textcolor{red}{a}}{\textcolor{b l u e}{b}}\right) = \frac{\left(\textcolor{red}{a}\right) ' \cdot \left(\textcolor{b l u e}{b}\right) - \left(\textcolor{b l u e}{b}\right) ' \cdot \left(\textcolor{red}{a}\right)}{\textcolor{b l u e}{b}} ^ 2$

Here, we can substitute:

d/dx ((color(red)(t^2))' * ((color(blue)(t-4))(color(blue)(2-t^3))- ((color(blue)(t-4)) * (color(blue)(2-t^3)))' * (color(red)(t^2)))) / (((color(blue)(2-t^3)) * (color(blue)(t-4)))^2

Derivative of $\textcolor{red}{{t}^{2}} = \textcolor{red}{2 t}$

Derivative of $\textcolor{b l u e}{\left(t - 4\right) \left(2 - {t}^{3}\right)}$ is a little complicated. We must use Product Rule , which states:

$\frac{d}{\mathrm{dx}} \left(\textcolor{red}{a} \cdot \textcolor{b l u e}{b}\right) = \left(\textcolor{red}{a}\right) ' \cdot \left(\textcolor{b l u e}{b}\right) + \left(\textcolor{b l u e}{b}\right) ' \cdot \left(\textcolor{red}{a}\right)$

We can substitute:

$\to \left(\textcolor{red}{t - 4}\right) ' \cdot \left(\textcolor{b l u e}{2 - {t}^{3}}\right) + \left(\textcolor{red}{t - 4}\right) \cdot \left(\textcolor{b l u e}{2 - {t}^{3}}\right) '$

$= \textcolor{red}{1} \cdot \left(\textcolor{b l u e}{2 - {t}^{3}}\right) + \left(\textcolor{red}{t - 4}\right) \cdot \left(\textcolor{b l u e}{3 {t}^{2}}\right)$

$= \textcolor{b l u e}{2 - {t}^{3} + 3 {t}^{3} - 12 {t}^{2}}$

$\frac{d}{\mathrm{dx}} \left(\textcolor{b l u e}{t - 4}\right) \cdot \left(\textcolor{b l u e}{2 - {t}^{3}}\right) = \textcolor{b l u e}{2 {t}^{3} - 12 {t}^{2} + 2}$

Now, we can substitute everything into our first equation:

-> ((color(red)(2t)) * (color(blue)(t-4))(color(blue)(2-t^3)) - (color(blue)(2t^3-12t^2+2)) * (color(red)(t^2)))/(((color(blue)(2-t^3)) * (color(blue)(t-4)))^2

Simplifying:

((color(red)(−2t^5+8t^4+4t^2−16t)) - (color(blue)(2t^5−12t^4+2t^2)))/((color(blue)(2-t^3))^2(color(blue)(t-4))^2

$= \frac{2 {t}^{5} - 4 {t}^{4} + 2 {t}^{2} - 16 t}{{\left(2 - {t}^{3}\right)}^{2} \cdot {\left(t - 4\right)}^{2}}$

Thus, solved.