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

Question

3187 views around the world

You can reuse this answer
Creative Commons License

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.