1 Answer

Yogur · Answer 1 · 2021-11-18T09:04:45+0000

Assuming a is constant and you want the derivatives with respect to t, we have by the chain rule

$(a\cos^3t)'=3a\cos^2t(\cos t)'=-3a\cos^2t\sin t$

$(a\sin^3t)'=3a\sin^2t(\sin t)'=3a\sin^2t\cos t$

Then for the second derivatives, we use the chain and product rules together:

$(a\cos^3t)''=(-3a\cos^2t\sin t)'=-6a\cos t\sin t(\cos t)'-3a\cos^3t$

$(a\cos^3t)''=3a\cos t(2\sin^2t-\cos^2t)$

$(a\sin^3t)''=(3a\sin^2t\cos t)'=6a\sin t\cos t(\sin t)'-3a\sin^3t$

$(a\sin^3t)''=3a\sin t(2\cos^2t-\sin^2t)$

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