1. n=1, differentiate y, y' = e^x + y,
n=2, y'' = e^x + y'
...
therefore d^(n)y/dx^n = ne^x + xe^x
2. apparently true for n=1
using the formula sinαsinβ = [cos(α-β) - cos(α+β)] / 2
n=2, (sinx + sin2x)sin(x/2) = [cos(x/2) - cos(3x/2)] /2 + [cos(3x/2) - cos(5x/2)] /2 = [cos(x/2) - cos(5x/2)] /2...