# What is dy/dx from first principle for y=3x^2-2x ?

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### 1 Answer

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

**- 2h -**

*- 2x*

*3x^2***)/h**

*+ 2x*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.