let ( h(t)=4 t^{3.2}-3 t^{-3.2} ). compute the following.\n( h^{prime}(t)= )\n( h^{prime}(1)= )\n( h^{prime…

let ( h(t)=4 t^{3.2}-3 t^{-3.2} ). compute the following.\n( h^{prime}(t)= )\n( h^{prime}(1)= )\n( h^{prime prime}(t)= )\n( h^{prime prime}(1)= )\nnote: you can earn partial credit on this problem.\nnote: you are in the reduced scoring period. all work counts for ( 85 % )\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 5 attempts remaining.

let ( h(t)=4 t^{3.2}-3 t^{-3.2} ). compute the following.\n( h^{prime}(t)= )\n( h^{prime}(1)= )\n( h^{prime prime}(t)= )\n( h^{prime prime}(1)= )\nnote: you can earn partial credit on this problem.\nnote: you are in the reduced scoring period. all work counts for ( 85 % )\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 5 attempts remaining.

Answer

Explanation:

Step1: Differentiate (h(t)) using power rule

The power rule is ((x^n)^\prime = nx^{n - 1}). For (y = 4t^{3.2}-3t^{-3.2}), (h^\prime(t)=(4t^{3.2})^\prime-(3t^{-3.2})^\prime). ((4t^{3.2})^\prime = 4\times3.2t^{3.2 - 1}=12.8t^{2.2}), ((3t^{-3.2})^\prime=3\times(- 3.2)t^{-3.2 - 1}=-9.6t^{-4.2}). So (h^\prime(t)=12.8t^{2.2}+9.6t^{-4.2}).

Step2: Calculate (h^\prime(1))

Substitute (t = 1) into (h^\prime(t)). Since (t^{2.2}=1^{2.2}=1) and (t^{-4.2}=1^{-4.2}=1), then (h^\prime(1)=12.8\times1+9.6\times1 = 22.4).

Step3: Differentiate (h^\prime(t)) to get (h^{\prime\prime}(t))

Using the power rule again. ((12.8t^{2.2})^\prime=12.8\times2.2t^{2.2 - 1}=28.16t^{1.2}), ((9.6t^{-4.2})^\prime=9.6\times(-4.2)t^{-4.2 - 1}=-40.32t^{-5.2}). So (h^{\prime\prime}(t)=28.16t^{1.2}+40.32t^{-5.2}).

Step4: Calculate (h^{\prime\prime}(1))

Substitute (t = 1) into (h^{\prime\prime}(t)). Since (t^{1.2}=1^{1.2}=1) and (t^{-5.2}=1^{-5.2}=1), then (h^{\prime\prime}(1)=28.16\times1+40.32\times1=68.48).

Answer:

(h^\prime(t)=12.8t^{2.2}+9.6t^{-4.2}); (h^\prime(1)=22.4); (h^{\prime\prime}(t)=28.16t^{1.2}+40.32t^{-5.2}); (h^{\prime\prime}(1)=68.48)