Find dy/dx from first principles if y=2x^2?

dy/dx = lim [f(x+h) - f(x)]/h, if h approaches to
0.

lim [f(x+h) - f(x)]/h = lim [2(x+h)^2 -
2x^2]/h

We'll expand the
binomial:

lim [2(x+h)^2 - 2x^2]/h = lim (2x^2 + 4xh + 2h^2
- 2x^2)/h

We'll eliminate like terms inside
brackets:

lim (2x^2 + 4xh + 2h^2 - 2x^2)/h = lim (4xh +
2h^2)/h

We'll factorize by 2h the
numerator:

lim (4xh + 2h^2)/h = lim 2h*(2x +
h)/h

We'll simplify and we'll
get:

lim 2h*(2x + h)/h = lim 2*(2x +
h)

We'll substitute h by the value of accumulation
point:

lim 2*(2x + h) = 4x +
2*0

lim 2*(2x + h) =
4x

The value of dy/dx from first principles
is: dy/dx = 4x.