What is dy/dx from first principle for y=3x^2-2x ?
We'll apply delta method to determine the instantaneous rate of change of y with respect to x.
dy/dx = lim [f(x + delta x) - f(x)]/delta x, delta x->0
We also can write:
dy/dx = lim [f(x + h) - f(x)]/h, h->0
We'll calculate f(x+h) = 3(x+h)^2 - 2(x+h)
We'll raise to square x + h:
f(x+h) = 3x^2 + 6xh + 3h^2 - 2x - 2h
dy/dx = lim (3x^2 + 6xh + 3h^2 - 2x - 2h - 3x^2 + 2x)/h
We'll eliminate like terms:
dy/dx = lim (6xh + 3h^2 - 2h)/h
lim (6xh + 3h^2 - 2h)/h = lim (6x + 3h - 2)
We'll substitute h by 0:
lim (6x + 3h - 2) = 6x - 2
dy/dx = 6x - 2
Substituting x by any value, we can compute the slope of the tangent to the graph of the function, in the chosen value for x.