Question

Find y' and y''. y = eαx sin βx

Answer 1

 Given: y = e^(αx)sin(βx) 
To find the first derivative, simply use the Product and Chain Rules: y' = e^(αx)cos(βx)(β) + sin(βx)(αe^(αx)) y' = αe^(αx)sin(βx) + βe^(αx)cos(βx) 
You could factor out the e^(αx) if you wanted. It will make it easier when taking the second derivative. y' = e^(αx)[αsin(βx) + βcos(βx)] 
To find the second derivative, simple take the derivative of y'. This one involves multiple Chain rules and multiple Product rules. y'' = e^(αx)[[(α)(cos(βx)(β) + sin(βx)(0)] + [(β)(-sin(βx)(β) + cos(βx)(0)]] + [αsin(βx) + βcos(βx)](αe^(αx)) 
That's a lot. Simplify: y'' = e^(αx)[[αβcos(βx)] + [-β²sin(βx)]] + [αsin(βx) + βcos(βx)](αe^(αx)) 
Distribute: y'' = αβe^(αx)cos(βx) − β²e^(αx)sin(βx) + α²e^(αx)sin(βx) + αβe^(αx)cos(βx) 
And finally, add like-terms and order: y'' = α²e^(αx)sin(βx) − β²e^(αx)sin(βx) + 2αβe^(αx)cos(βx)