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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...

It was part of the National Qualifying Exams held by ASO. How did you go in this year's AChO?