# What is the derivative of x^3 - x^2 + x + 1 from first principles?

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### 1 Answer

The derivative of a function f(x) from first principles is f'(x) = `lim_(h->0)[f(x + h) - f(x)]/h`

Here f(x) = x^3 - x^2 + x + 1

f'(x) = `lim_(h->0) [(x + h)^3 - (x + h)^2 + (x + h) + 1 - (x^3 - x^2 + x + 1)]/h`

=> f'(x) = `lim_(h->0)[x^3 + 3x^2h + 3xh^2 + h^3 - x^2 - h^2 - 2xh + x + h + 1 - x^3 + x^2 - x - 1]/h`

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

=> f'(x) = `lim_(h->0) 3x^2 + 3xh + h^2 - h - 2x + 1`

=> 3x^2 - 2x + 1

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