How do you find the first and second derivative of #y=e^(alphax)sinbetax#?

1 Answer
Feb 7, 2017

#y=e^(alphax)sin(betax)#

#=>#

# y''=alpha^2e^(alphax)sin(betax)+2alphabetae^(alphax)cos(betax)-beta^2e^(alphax)sin(betax)#

Explanation:

First, remember the product rule states

#(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)#

In this case

#f(x)=e^(alphax)#

and

#g(x)=sin(betax)#

In order to find the derivative of #y(x)#, using the product rule, we will first need to find #f'(x)# and #g'(x)#.

To find these, we will need the chain rule.

#(a(b(x))'=a'(b(x))b'(x)#

In the case of #f(x)#

#a(x)=e^x#

and

#b(x)=alphax#

Since #e^x# is its own derivative

#f'(x)=(b'(x))e^(b(x))=(alphax)'e^(alphax)#

And since #(cx)'=c(x)'=c#

#f'(x)=alphae^(alphax)#

and in the case of #g(x)#

#a(x)=sin(x)#

and

#b(x)=betax#

Then, using the same rules from above we have

#g'(x)=(sin(betax))'=(betax)'sin'(betax)=betasin'(betax)#

and since #(sin(x))'=cos(x)#

#g'(x)=betacos(betax)#

Then plugging in we get

#y=e^(alphax)sin(betax)=> underline(y'=alphae^(alphax)sin(betax)+betae^(alphax)cos(betax))#

Now, we have the first derivative. So, to find the second derivative, we now take the derivative of #y'(x)#.

Since #(f(x)+g(x))'=f'(x)+g'(x)#

Then

#y''=(alphae^(alphax)sin(betax))'+(betae^(alphax)cos(betax))'#

By above

#(alphae^(alphax)sin(betax))'=alpha(e^(alphax)sin(betax))'=#

#alpha(alphae^(alphax)sin(betax)+betae^(alphax)cos(betax))#

#=alpha^2e^(alphax)sin(betax)+alphabetae^(alphax)cos(betax)#

also by above

#(betae^(alphax)cos(betax))'=beta(e^(alphax)cos(betax))'#

#=beta(alphae^(alphax)cos(betax)+e^(alphax)cos'(betax))#

Since #(cos(x))'=-sin(x)#

#=beta(alphae^(alphax)cos(betax)-betae^(alphax)sin(betax))#

#=alphabetae^(alphax)cos(betax)-beta^2e^(alphax)sin(betax)#

Then we plug in

#y''=alpha^2e^(alphax)sin(betax)+alphabetae^(alphax)cos(betax)+alphabetae^(alphax)cos(betax)-beta^2e^(alphax)sin(betax)=underline(alpha^2e^(alphax)sin(betax)+2alphabetae^(alphax)cos(betax)-beta^2e^(alphax)sin(betax))#