Question

How do you find the derivative of `4x^3-3x+8` by first principles?

Answer 1

`d/(dx) (4x^3-3x+8) = 12x^2-3`

Explanation:

Given two points `(x_1, y_1)` and `(x_2, y_2)`, the slope `m` of a line through those points is given by the formula:

`m = (Delta y)/(Delta x) = (y_2-y_1)/(x_2-x_1)`

If those two points are `(x, f(x))` and `(x+h, f(x+h))` then we find that the slope of the secant joining those points is:

`m = (f(x+h)-f(x))/((x+h)-x) = (f(x+h)-f(x))/h`

As `h->0` the second point approaches the first and if the limit exists, then it gives the instantaneous slope at the point `(x, f(x))`, otherwise known as the derivative.

Given:

`f(x) = 4x^3-3x+8`

We have:

`(f(x+h)-f(x))/h`

`= ((4(x+h)^3-3(x+h)+8)-(4x^3-3x+8))/h`

`= ((4(x^3+3hx^2+3h^2x+h^3)-3(x+h)+8)-(4x^3-3x+8))/h`

`= ((color(red)(cancel(color(black)(4x^3)))+12hx^2+12h^2x+4h^3-color(red)(cancel(color(black)(3x)))-3h+color(red)(cancel(color(black)(8))))-(color(red)(cancel(color(black)(4x^3)))-color(red)(cancel(color(black)(3x)))+color(red)(cancel(color(black)(8)))))/h`

`= (12hx^2+12h^2x+4h^3-3h)/h`

`= 12x^2+12hx+4h^2-3`

So:

`d/(dx) f(x) = lim_(h->0) (f(x+h)-f(x))/h`

`color(white)(d/(dx) f(x)) = lim_(h->0) 12x^2+12hx+4h^2-3`

`color(white)(d/(dx) f(x)) = 12x^2-3`