How do you differentiate h(t)=(t^4-1)^3(t^3+1)^-2?

Oct 20, 2016

$h ' \left(t\right) = - 2 {\left({t}^{4} - 1\right)}^{3} / {\left({t}^{3} + 1\right)}^{3} + 12 {t}^{3} {\left({t}^{4} - 1\right)}^{2} / {\left({t}^{3} + 1\right)}^{2}$

Explanation:

You must use a combination of the chain rule and product rule

product rule

$h ' \left(t\right) = u v ' + u ' v$

$u = {\left({t}^{4} - 1\right)}^{3}$

$u ' = 3 {\left({t}^{4} - 1\right)}^{2} \cdot 4 {t}^{3} \implies$ Apply the chain rule find the derivative of ${t}^{4} - 1$

$v = {\left({t}^{3} + 1\right)}^{- 2}$

$v ' = - 2 {\left({t}^{3} + 1\right)}^{- 3} \implies$ Apply the chain rule find the derivative of ${t}^{3} + 1$

Put everything together using substitution

$h ' \left(t\right) = {\left({t}^{4} - 1\right)}^{3} \cdot \left(- 2\right) {\left({t}^{3} + 1\right)}^{- 3} + 3 {\left({t}^{4} - 1\right)}^{2} \cdot 4 {t}^{3} {\left({t}^{3} + 1\right)}^{- 2}$

Simplify

$h ' \left(t\right) = - 2 {\left({t}^{4} - 1\right)}^{3} / {\left({t}^{3} + 1\right)}^{3} + 12 {t}^{3} {\left({t}^{4} - 1\right)}^{2} / {\left({t}^{3} + 1\right)}^{2}$

Here is another example to show the usage of the chain and power rules.