Calculus: Early Transcendentals, Chapter 3, 3.4, Section 3.4, Problem 49

Using the product rule:
d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x)
y'= d/dx [e^(alpha x) sin (beta x)]
= e^(alpha x) * alpha * sin(beta x) + e^(alpha x) * cos(beta x)* beta
= e^(alpha x) *[alpha sin(beta x) + beta * cos (beta x)]
y'' = d/dx (e^(alpha x)) * [alpha sin (beta x) + beta cos (beta x)]
+ e^(alpha x) d/dx [alpha sin(betax) + beta cos(betax)]
= alpha e^(alphax) * [alpha sin(beta x)+ beta cos(betax)]
+e^(alpha x)*[alpha*beta cos(beta x)-beta^2 sin(betax)]
= e^(alpha x) *[(alpha^2-beta^2)sin(betax) + 2alphabeta cos(betax)]
hope this helps.