1. ## 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) + e–x
y' = ?
y'' = ?

2. 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!

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 $\displaystyle y''$

3. 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 $\displaystyle y'' = -27 sin(3x) + e^{-x}$

4. 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.

5. Yeah, I edited, lol. I went too quick!

Thanks, man!