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Section 3.12 : Higher Order Derivatives
6. Determine the second derivative of \(g\left( x \right) = \sin \left( {2{x^3} - 9x} \right)\)
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Start SolutionNot 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,
\[g'\left( x \right) = \left( {6{x^2} - 9} \right)\cos \left( {2{x^3} - 9x} \right)\] Show Step 2Do 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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{g''\left( x \right) = 12x\cos \left( {2{x^3} - 9x} \right) - {{\left( {6{x^2} - 9} \right)}^2}\sin \left( {2{x^3} - 9x} \right)}}\]