1 maths Q (What's wrong?)

y = x^2 /2
y+δy = (x=+δx)^2 /2
x^2 /2 + δy = x^2/2 + xδx + (δx)^2 /2
δy = xδx + (δx)^2 /2

Since δx is assumed to tend to 0, then it turns to differentiation, i.e. δx -> 0
=> dy = xdx + 0, which is done in differentiation



2012-01-29 17:25:22 補充：
p.s. It's better for you to write dy/dx = x since it's a much more formal expression.

2012-01-29 17:36:03 補充：
p.s. It's not dx^2 /2 but δx^2 /2
Since δx -> 0, then δx^2 /2 -> 0^2/2 = 0

2012-01-29 17:41:11 補充：
Actually, dx^2 /2 is reducdant since δx^2 /2 -> 0^2/2 = 0 already, we need not to state the appearance of dx^2 /2