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))#
