# How do you use the quotient rule to differentiate y =((3t+4)/(6t+7))^3?

Jul 17, 2016

$= - 9 {\left(3 t + 4\right)}^{2} / {\left(6 t + 7\right)}^{4}$

#### Explanation:

$y = {\left(\frac{3 t + 4}{6 t + 7}\right)}^{3}$

a spot of chain rule first to set it up

$\frac{d}{\mathrm{dt}} {\left(\frac{3 t + 4}{6 t + 7}\right)}^{3} = 3 {\left(\frac{3 t + 4}{6 t + 7}\right)}^{2} \frac{d}{\mathrm{dt}} \left(\frac{3 t + 4}{6 t + 7}\right)$

and that's were the quotient rule comes in $\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{v} ^ 2$

$\frac{d}{\mathrm{dt}} \left(\frac{3 t + 4}{6 t + 7}\right) = \frac{3 \cdot \left(6 t + 7\right) - \left(3 t + 4\right) \cdot 6}{6 t + 7} ^ 2$

$= - \frac{3}{6 t + 7} ^ 2$

putting it all together

$\frac{d}{\mathrm{dt}} {\left(\frac{3 t + 4}{6 t + 7}\right)}^{3} = 3 {\left(\frac{3 t + 4}{6 t + 7}\right)}^{2} \cdot - \frac{3}{6 t + 7} ^ 2$

$= - 9 {\left(3 t + 4\right)}^{2} / {\left(6 t + 7\right)}^{4}$