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Math Help - Second Derivative

  1. #1
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    Second Derivative

    How would I go about finding the second derivative to this equation? This is a homework problem and it's really confusing me, any help would be appreciated!

    y = 3sin(3x) + ex
    y' = ?
    y'' = ?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by KennyP33 View Post
    How would I go about finding the second derivative to this equation? This is a homework problem and it's really confusing me, any help would be appreciated!

    y = 3sin(3x) + e–x
    y' = ?
    y'' = ?
    I assume that you mean y=3\sin{3x}+e^{-x}

    So Recall: \frac{d}{dx}\sin{u}=u'\cos{u} and \frac{d}{dx}e^{u}=e^uu'.

    So, y'=3(3)\cos(3x)+(-1)e^{-x}.

    Differentiate again to find y''
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    I assume that you mean y=3\sin{3x}+e^{-x}

    So Recall: \frac{d}{dx}\sin{u}=u'\cos{u} and \frac{d}{dx}e^{u}=e^uu'.

    So, y'=3(3)\cos(3x)+(-1)e^{-x}.

    Differentiate again to find y''
    Thank you!

    so would the 2nd derivation be y'' = -27 sin(3x) + e^{-x}
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by KennyP33 View Post
    Thank you!

    so would the 2nd derivation be y'' = -27 sin(3x) + {\color{red}1}e^{-x}
    Good except for the correction that I made.

    EDIT: Oops, I see that you caught it. Good work.
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  5. #5
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    Yeah, I edited, lol. I went too quick!

    Thanks, man!
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