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!
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 $\displaystyle \frac{d}{dx}e^{u}=e^uu'$.