Derivar $$ \text { 1. } f(x)=3 x^{2}-3 x-1 $$

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Alright, I need to find the derivative of the function $$ f(x) = 3x^2 - 3x - 1 $$. Let's start by recalling the basic rules of differentiation. The derivative of a function gives the rate at which the function's value changes with respect to its variable, in this case, $$ x $$.

First, I'll consider each term in the function separately. The function is a polynomial, so I can differentiate each term individually and then sum the results.

1. **Differentiating $$ 3x^2 $$:**
   - The power rule of differentiation states that $$ \frac{d}{dx}x^n = n x^{n-1} $$.
   - Applying this to $$ 3x^2 $$, the derivative is $$ 3 \times 2x^{2-1} = 6x $$.

2. **Differentiating $$ -3x $$:**
   - The derivative of $$ x $$ with respect to $$ x $$ is 1.
   - Therefore, the derivative of $$ -3x $$ is $$ -3 \times 1 = -3 $$.

3. **Differentiating $$ -1 $$:**
   - The derivative of a constant is always 0, since constants do not change as $$ x $$ changes.
   - So, the derivative of $$ -1 $$ is 0.

Now, combining the derivatives of all terms:
$$ f'(x) = 6x - 3 + 0 = 6x - 3 $$

But wait, let me double-check to ensure I haven't missed anything. Another way to approach this is to use the limit definition of the derivative:
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$

Let's compute $$ f(x+h) $$:
$$ f(x+h) = 3(x+h)^2 - 3(x+h) - 1 $$
$$ = 3(x^2 + 2xh + h^2) - 3x - 3h - 1 $$
$$ = 3x^2 + 6xh + 3h^2 - 3x - 3h - 1 $$

Now, subtract $$ f(x) $$ from $$ f(x+h) $$:
$$ f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 3x - 3h - 1) - (3x^2 - 3x - 1) $$
$$ = 6xh + 3h^2 - 3h $$

Divide by $$ h $$:
$$ \frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 - 3h}{h} $$
$$ = 6x + 3h - 3 $$

Take the limit as $$ h \to 0 $$:
$$ f'(x) = 6x - 3 $$

This confirms my earlier result. Therefore, the derivative of $$ f(x) = 3x^2 - 3x - 1 $$ is indeed $$ f'(x) = 6x - 3 $$.

La derivada de la función $$ f(x) = 3x^2 - 3x - 1 $$ es $$ f'(x) = 6x - 3 $$.

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