Question

Using the limit definition, how do you find the derivative of `f(x) = 3x^2-5x+2`?

Answer 1

See explanation.

Explanation:

The derivative at `x_0` is:

`f'(x_0)=lim_{h->0}(f(x_0+h)-f(x_0))/h`

Using this definition we get:

`f'(x_0)=lim_{h->0}(f(x_0+h)-f(x_0))/h=lim_{h->0}([3(x_0+h)^2-5(x_0+h)+2]-[3x_0^2-5x_0+2])/h`

`f'(x_0)=lim_{h->0}(3(x_0+h)^2-5x_0-5h+2-3x_0^2+5x_0-2)/h=`

`=lim_{h->0}(3x_0^2+6x_0h+h^2-5x_0-5h+2-3x_0^2+5x_0-2)/h=`

`=lim_{h->0}(6x_0h-5h+h^2)/h=lim_{h->0}(h*(6x_0-5+h))/h=`

`=lim_{h->0}(6x_0-5+h)=6x_0-5`

Finally we can write that `f'(x_0)=6x_0-5`

QED