# y = 3x^2-x Find the derivative using first principles.

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### 2 Answers

The derivative of a function f(x) from first principles is:

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

For f(x) = 3x^2 - x

`f'(x) = lim_(h->0) (3(x + h)^2 - (x + h) - (3x^2 - x))/h`

= `lim_(h->0) (3x^2 + 6xh + 3h^2 - x - h - 3x^2 + x)/h`

= `lim_(h->0) (6xh + 3h^2 - h)/h`

= `lim_(h->0) (6x + 3h - 1)`

Substituting h = 0

=> 6x - 1

**The derivative of f(x) = 3x^2 - x is f'(x) = 6x - 1**

`y=3x^2-x`

`y=3x^2-x` ------------(1)

For `deltax` small change in x if the change in y is ` deltay ` is then;

`y+deltay=3(x+deltax)^2-(x+deltax)` ---------------(2)

(2)-(1)

`deltay=3(x^2+2x*deltax+deltax ^2)-(x+deltax)-(3x^2-x)`

`deltay =6x*deltax-deltax-3deltax ^2`

Since `deltax` is very small `deltax^2=0`

`deltay =6x*deltax-deltax = deltax(6x-1)`

`(dy)/dx = lim_(deltaxrarr0) (deltay)/(deltax)`

`(dy)/dx = lim_(deltaxrarr0)(deltax(6x-1))/(deltax)`

`(dy)/dx =6x^-1`

*So the derivative of y=3x^2-x by first principles is 6x-1.*

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