Question

How do you use the definition of a derivative to find the derivative of `f(x)=x^3-2x^2+5x-6`?

Answer 1

`f'(x)=3x^2-4x+5`

Explanation:

Here ,

`f(x)=x^3-2x^2+5x-6=>f(t)=t^3-2t^2+5t-6`

We know that ,

`f'(x)=lim_(t tox)(f(t)-f(x))/(t-x) to"definition"`

Substitute values of `f(t) and f(x)`

`f'(x)=lim_(t tox)((t^3-2t^2+5t-6)-(x^3-2x^2+5x-6))/(t-x)`

`f'(x)=lim_(t tox)((t^3-x^3)-2(t^2-x^2)+5(t-x))/(t-x)`

`f'(x)=lim_(t tox)(color(red)(cancel((t-x))){(t^2+tx+x^2)-2(t+x)+5(1)})/color(red)cancel(t-x)`

`f'(x)=lim_(t tox){(t^2+tx+x^2)-2(t+x)+5(1)}`

`f'(x)=(x^2+x*x+x^2)-2(x+x)+5(1)`

`f'(x)=(3x^2)-2(2x)+5(1)`

`:.f'(x)=3x^2-4x+5`

Answer 2

`f'(x) = 3 x^2 - 4 x + 5`

Explanation:

`"The definition of the derivative of "f(x)" is : "`

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

`"So here we have "`

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

`= ((x^3 + 3 h x^2 + 3 h^2 x + h^3) - 2 (x^2 + 2 h x + h^2) + 5 (x + h) - x^3 + 2 x^2 - 5 x)/h`

`= lim_{h->0} (3 h x^2 + 3 h^2 x + h^3 - 4 h x - 2 h^2 + 5 h)/h`

`= lim_{h->0} 3 x^2 + 3 h x + h^2 - 4 x - 2 h + 5`

`= lim_{h->0} 3 x^2 - 4 x + 5 + h^2 + 3 h x - 2 h`

`= 3 x^2 - 4 x + 5 + lim_{h->0} h^2 + 3 h x - 2 h`

`= 3 x^2 - 4 x + 5 + 0`

`= 3 x^2 - 4 x + 5`