y = e^xsinx y'' = e^xcosx - e^xsinx correct? much appreciated.
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maybe not.. what is your first derivative?
Originally Posted by kalagota maybe not.. what is your first derivative? the first derivative is y' = e^xsinx + e^xcosx
Originally Posted by jvignacio the first derivative is y' = e^xsinx + e^xcosx Correct, and what is the second derivative? (Note your answer in your first post is not correct.)
Originally Posted by Chop Suey Correct, and what is the second derivative? (Note your answer in your first post is not correct.) hmm is it. y'' = (e^xsinx + e^xcosx) + (e^xcosx - e^xsinx) ??
yes indeed! and you should see that the terms with sin x cancel..
Originally Posted by kalagota yes indeed! and you should see that the terms with sin x cancel.. so it leaves e^xcosx + e^xcosx ??
yes.. or simply $\displaystyle 2e^x\cos x$
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