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For the derivative of $5+ 10e^{-t}\sin(2t-30)$ I am getting this result:

$$ -20e^{-t}\sin(2t-30) + 10e^{-t}\cos t2t, $$

BUT my textbook says the answer is:

$$ 22.36e^{-t}\sin(2t+86.565). $$

Could someone explain how this result is produced?

mookid
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AOE
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    Why the tag "Laplace-transform"? –  Mar 13 '14 at 02:25
  • the question was asking about the impulse response from a step response which have to do with laplace transforms since the impulse response is the inverse laplace of the transfer function – AOE Mar 13 '14 at 02:53
  • Oh. It would be good to add that in the question, so that others understand why the tag was used. –  Mar 13 '14 at 02:54

1 Answers1

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By your derivative, I understand that you mean $$f=5+10e^{-t}\sin(2t-30)$$If so, then by the product rule,$$f=5+10e^{-t}\sin(2t-30)\\ f^{'}=10(-\sin(2t-30)e^{-t}+2e^{-t}\cos(2t-30))$$Now, all you have to do is use the identity of $\sin(a+b)=\sin a\cos b+\sin b\cos a,\cos(a+b)=\cos a\cos b-\sin a \sin b$ to get the textbook's result.