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Section 3.12 : Higher Order Derivatives

9. Determine the second derivative of \(H\left( t \right) = {\cos ^2}\left( {7t} \right)\)

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Not much to this problem other than to take two derivatives so each step will show each successive derivative until we get to the second. The first derivative is then,

\[H'\left( t \right) = - 14\cos \left( {7t} \right)\sin \left( {7t} \right)\] Show Step 2

Do not forget that often we will end up needing to do a product rule in the second derivative even though we did not need to do that in the first derivative. The second derivative is then,

\[H''\left( t \right) = 98\sin \left( {7t} \right)\sin \left( {7t} \right) - 98\cos \left( {7t} \right)\cos \left( {7t} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{98{{\sin }^2}\left( {7t} \right) - 98{{\cos }^2}\left( {7t} \right)}}\]

Note that, in this case, if we recall our trig formulas we could have reduced the product in the first derivative to a single trig function which would have then allowed us to avoid the product rule for the second derivative. Can you figure out what the formula is?