derivative of h(t)=(t+1)^(2/3)(2t^2-1)^3
Are you familiar with the product rule?
if y = ab where a and b are functions of t and y = h(t), then h'(t) = a'b + ab'.
You will also end up using the chain rule once you break the problem up into a'b + ab'.
Making sense to you?
There is a hierarchy of rule application here. First the product rule and then the chain rule once you have the problem separated by the product rule.
the answer is h'(t)=2/3(t+1)^(-1/3)(2t^2-1)^2(20t^2+18t-1) what i don't get is the last part (20t^2+18t-1)
That comes about by factoring the answer. Once you get an intermediate answer, there is a factor of (t+1)^(-1/3) and (2t^2-1)^2 in both parts. They are just recombining as much as possible.
o now i get thanks
You're welcome and good luck with your studies.
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