second dervative of Asin(mx)+Bcos(nx)

Question

second dervative of Asin(mx)+Bcos(nx)

Answer 1

A,B,m,n are all constants, only need to use the chain rule.

Know your trig derivatives:

d(sinx)/dx = cosx

d(cosx)/dx = -sinx

 

f(x) = A*sin(mx)+B*cos(nx)

f'(x) = Am*cos(mx) - Bn*sin(nx)

f''(x) = -Am²*sin(mx) - Bn²*cos(mx)