You also may use `Delta` method to find derivative from first principle, such that:

`lim_(Delta x->0) (f(Delta x + h) - f(x))/(Delta x)`

Replacing `2(x+Delta x)^2+3(x+Delta x)` for `f(x + h)` and `2x^2+3x` for `f(x)` , yields:

`lim_(Delta x->0) (2(x+Delta x)^2 + 3(x+Delta x) - 2x^2 - 3x)/(Delta x)`

`lim_(Delta x->0) (2x^2 + 4xDelta x + 2(Delta x)^2 + 3x + 3Delta x - 2x^2 - 3x)/(Delta x)`

Reducing duplicate terms yields:

`lim_(Delta x->0) (4xDelta x + 2(Delta x)^2 + 3h)/(Delta x)`

Factoring out h yields:

`lim_(Delta x->0) Delta x(4x + 2Delta x + 3)/(Delta x) `

Reducing duplicate factors yields:

`lim_(Delta x->0) (4x + 2Delta x + 3) = 4x + 0 + 3 `

**Hence, evaluating derivative of the function from the first principle, using `Delta` method, yields **`lim_(Delta x->0) (2(x+Delta x)^2 + 3(x+Delta x) - 2x^2 - 3x)/(Delta x) = 4x + 3.`

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) = 2(x+h)^2 + 3(x+h)

We'll raise to square x + h:

f(x+h) = 2x^2 + 4xh + 2h^2 + 3x + 3h

dy/dx = lim (2x^2 + 4xh + 2h^2 + 3x + 3h - 2x^2 - 3x)/h

We'll eliminate like terms:

dy/dx = lim (4xh + 2h^2 + 3h)/h

lim (4xh + 2h^2 + 3h)/h = lim (4x + 2h + 3)

We'll substitute h by 0:

lim (4x + 2h + 3) = 4x + 3

dy/dx = 4x + 3

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.