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

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Answer 1 · 2016-10-20T21:38:54

$h'(t)=-2(t^4-1)^3/(t^3+1)^(3)+12t^3(t^4-1)^2/(t^3+1)^(2)$

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

product rule

$h'(t)=uv'+u'v$

$u=(t^4-1)^3$

$u'=3(t^4-1)^2*4t^3 =>$ Apply the chain rule find the derivative of $t^4-1$

$v=(t^3+1)^(-2)$

$v'=-2(t^3+1)^(-3) =>$ Apply the chain rule find the derivative of $t^3+1$

Put everything together using substitution

$h'(t)=(t^4-1)^3*(-2)(t^3+1)^(-3)+3(t^4-1)^2*4t^3(t^3+1)^(-2)$

Simplify

$h'(t)=-2(t^4-1)^3/(t^3+1)^(3)+12t^3(t^4-1)^2/(t^3+1)^(2)$

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

How do you differentiate h(t)=(t^4-1)^3(t^3+1)^-2h(t)=(t4−1)3(t3+1)−2h(t)=(t^4-1)^3(t^3+1)^-2?