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

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 \( f'(x) = 6x - 3 \).