Question

let ( h(t)=4 t^{3.2}-3 t^{-3.2} ). compute the following.
( h^{prime}(t)= )
( h^{prime}(1)= )
( h^{prime prime}(t)= )
( h^{prime prime}(1)= )
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note: you are in the reduced scoring period. all work counts for ( 85 % )
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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:

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