step 1\nthe function (h(t)=(t^{4}-1)^{3}(t^{3}+1)^{6}) is a product, and so we must use the product rule to…

step 1\nthe function (h(t)=(t^{4}-1)^{3}(t^{3}+1)^{6}) is a product, and so we must use the product rule to find its derivative. also, the factors of the product are compositions, so finding their derivatives will require using the chain rule.\nusing the chain rule, the derivative of ((t^{4}-1)^{3}) is\n3(t^{4}-1)^{2}(4cdot t^{3})\nstep 2\nsimilarly, the derivative of ((t^{3}+1)^{6}) is\n6(t^{3}+1)^{5}(3cdot t^{2})\nstep 3\nusing these derivatives in the product rule, we get\nh(t)=(t^{4}-1)^{3}(6(t^{3}+1)^{5}(3cdot t^{2}))+(t^{3}+1)^{6}(3(t^{4}-1)^{2}(4cdot t^{3}))\nafter factoring and simplifying, we can conclude that the derivative is as follows.\nh(t)=\n

step 1\nthe function (h(t)=(t^{4}-1)^{3}(t^{3}+1)^{6}) is a product, and so we must use the product rule to find its derivative. also, the factors of the product are compositions, so finding their derivatives will require using the chain rule.\nusing the chain rule, the derivative of ((t^{4}-1)^{3}) is\n3(t^{4}-1)^{2}(4cdot t^{3})\nstep 2\nsimilarly, the derivative of ((t^{3}+1)^{6}) is\n6(t^{3}+1)^{5}(3cdot t^{2})\nstep 3\nusing these derivatives in the product rule, we get\nh(t)=(t^{4}-1)^{3}(6(t^{3}+1)^{5}(3cdot t^{2}))+(t^{3}+1)^{6}(3(t^{4}-1)^{2}(4cdot t^{3}))\nafter factoring and simplifying, we can conclude that the derivative is as follows.\nh(t)=\n

Answer

Explanation:

Step1: Apply chain - rule to ((t^{4}-1)^{3})

The chain - rule states that if (y = u^{n}) and (u = g(t)), then (y^\prime=nu^{n - 1}g^\prime(t)). For (y=(t^{4}-1)^{3}), let (u = t^{4}-1) and (n = 3). Then (y^\prime=3(t^{4}-1)^{2}\cdot4t^{3}).

Step2: Apply chain - rule to ((t^{3}+1)^{6})

For (y=(t^{3}+1)^{6}), let (u = t^{3}+1) and (n = 6). Then (y^\prime=6(t^{3}+1)^{5}\cdot3t^{2}).

Step3: Apply product - rule

The product - rule states that if (h(t)=f(t)g(t)), then (h^\prime(t)=f(t)g^\prime(t)+g(t)f^\prime(t)). Here, (f(t)=(t^{4}-1)^{3}) and (g(t)=(t^{3}+1)^{6}). [ \begin{align*} h^\prime(t)&=(t^{4}-1)^{3}\left(6(t^{3}+1)^{5}(3t^{2})\right)+(t^{3}+1)^{6}\left(3(t^{4}-1)^{2}(4t^{3})\right)\ &=18t^{2}(t^{4}-1)^{3}(t^{3}+1)^{5}+12t^{3}(t^{3}+1)^{6}(t^{4}-1)^{2}\ &=6t^{2}(t^{4}-1)^{2}(t^{3}+1)^{5}\left[3(t^{4}-1)+2t(t^{3}+1)\right]\ &=6t^{2}(t^{4}-1)^{2}(t^{3}+1)^{5}(3t^{4}-3 + 2t^{4}+2t)\ &=6t^{2}(t^{4}-1)^{2}(t^{3}+1)^{5}(5t^{4}+2t - 3) \end{align*} ]

Answer:

(h^\prime(t)=6t^{2}(t^{4}-1)^{2}(t^{3}+1)^{5}(5t^{4}+2t - 3))