# Second Derivative

• Jan 30th 2010, 01:28 PM
KennyP33
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! (Hi)

y = 3sin(3x) + e–x
y' = ?
y'' = ?
• Jan 30th 2010, 01:33 PM
VonNemo19
Quote:

Originally Posted by KennyP33
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! (Hi)

y = 3sin(3x) + e–x
y' = ?
y'' = ?

I assume that you mean $\displaystyle y=3\sin{3x}+e^{-x}$

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

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

Differentiate again to find $\displaystyle y''$
• Jan 30th 2010, 01:37 PM
KennyP33
Quote:

Originally Posted by VonNemo19
I assume that you mean $\displaystyle y=3\sin{3x}+e^{-x}$

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

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

Differentiate again to find $\displaystyle y''$

Thank you!

so would the 2nd derivation be $\displaystyle y'' = -27 sin(3x) + e^{-x}$
• Jan 30th 2010, 01:40 PM
VonNemo19
Quote:

Originally Posted by KennyP33
Thank you!

so would the 2nd derivation be $\displaystyle y'' = -27 sin(3x) + {\color{red}1}e^{-x}$

Good except for the correction that I made.

EDIT: Oops, I see that you caught it. Good work.
• Jan 30th 2010, 01:41 PM
KennyP33
Yeah, I edited, lol. I went too quick!

Thanks, man!