y=2sin(3x+1)+e^2
y'= [2sin(3x+1)]' + [e^2]'
= 2cos(3x+1)*(3x+1)' + 0
= 2cos(3x+1)*3
= 6cos(3x+1)
y=ln[x/1+x^2] = lnx - ln(1+x^2)
y'= [ lnx - ln(1+x^2)]'
= [lnx]' - [ln(1+x^2)]'
= 1/x - (1+x^2)'/(1+x^2)
= 1/x - 2x/(1+x^2)
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