Question

How do you use the limit definition of the derivative to find the derivative of `f(x)=-4x^2-5x-2`?

Answer 1

See below.

Explanation:

We start by finding the gradient in much the same way as we do for any function. namely:

`(y_2-y_1)/(x_2-x_1)`

If we have a point `P` on a curve with coordinates `(x, f(x))`, then another point `Q` near `P` has coordinates `(x+deltax, f(x+delta x))`, where `delta x` is a small increment of `x`. Then gradient is:

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

And the derivative is:

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

From example:

`((-4(x+delta x)^2-5(x+delta x)-2)-(-4x^2-5x-2))/(x+delta x - x)`

simplifying

`(-4x^2-8xdelta x-4(deltax)^2-5x-5delta x-2+4x^2+5x+2)/(x+delta x - x)`

`->=(-8xdelta x-4(deltax)^2-5delta x)/(delta x )`

Cancelling `delta x`

`(-8x-4(deltax)-5)`

`d/dx=lim_(deltax->0)(-8x-4(deltax)-5)=-8x-5`