QA00

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course Math 279

been quite a while since i've done this! haha

QA 00My calculus is a bit rusty, it’s been about a year since my last calculus class. But here goes!

3 sin (4t + 2)

f’(x) = 12 cos (4t + 2)

f’’(x) = -48 sin (4t + 2)

2 cos^2 (3t + 2)

f’(x) = 4 cos (3t -1) * -sin (3t -1) (3) = -12sin(3t -1) cos (3t -1)

f’’(x) = 36 - 172(cos (3t-1))^2

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Good, but those (3t - 1) quantities should all be (3t + 2).

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A sin (omega * t + phi)

f’(x) = A cos (omega * t + phi) (omega)

= omega * A cos (omega * t + phi)

f’’(x) = omega^2 * A sin (omega * t + phi)

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Good, but this last one should be negative. Derivative of cos is - sin.

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e^(t^2 -1)

f’(x) = e^(t^2 - 1)(2t) = 2te^(t^2-1)

f’’(x) = 2te^(t^2-1) = 2t(2te^(t^2 -1) + 2(e^(t^2 - 1))

= 4t e^(t^2-1)+ 2 e^(t^2-1)

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4t ^2 e^(t^2-1)+ 2 e^(t^2-1) in that last step.

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You missed a couple of details, but your overall approach was good throughout. Check my notes.

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