1)y = cos(x+y) y'= -sin(x+y) (1+y') y'+ y'sin(x+y) = -sin(x+y) y'= -sin(x+y)/[1+sin(x+y)]
2)y = x + 1/lny^2 y'= 1 - (2yy'/y^2)/(lny^2)^2 y'{1+2/y(lny^2)^2}= 1 y'= 1/[1+2/y(lny^2)^2]
y'= 1/[1+2/[y(y - x)^2]
3) 2x^2-4y^2-1=0 4x - 8yy' = 0 y'= x/2y
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