(a) Using the first principles of differentiation, find the first derivative of $$ f(x)=-3 x^{2}+\frac{2}{x} $$ at $$ x=-1 $$.

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To find the first derivative of the function $$ f(x) = -3x^{2} + \frac{2}{x} $$ at $$ x = -1 $$ using the **first principles of differentiation**, follow these steps:

### Step 1: Recall the Definition of the Derivative
The derivative of $$ f(x) $$ at a point $$ x $$ is given by:
$$
f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
$$

### Step 2: Compute $$ f(x + h) $$
$$
f(x + h) = -3(x + h)^2 + \frac{2}{x + h}
$$
Expand and simplify:
$$
f(x + h) = -3(x^2 + 2xh + h^2) + \frac{2}{x + h} = -3x^2 - 6xh - 3h^2 + \frac{2}{x + h}
$$

### Step 3: Find $$ f(x + h) - f(x) $$
$$
f(x + h) - f(x) = (-3x^2 - 6xh - 3h^2 + \frac{2}{x + h}) - (-3x^2 + \frac{2}{x}) = -6xh - 3h^2 + \frac{2}{x + h} - \frac{2}{x}
$$

### Step 4: Divide by $$ h $$ and Take the Limit
$$
f'(x) = \lim_{h \to 0} \frac{-6xh - 3h^2 + \frac{2}{x + h} - \frac{2}{x}}{h}
$$
Simplify each term:
$$
f'(x) = \lim_{h \to 0} \left( -6x - 3h + \frac{-2h}{x(x + h)} \right)
$$
As $$ h \to 0 $$, the term $$ -3h $$ approaches 0, and the fraction simplifies:
$$
\frac{-2h}{x(x + h)} = \frac{-2}{x^2} \quad (\text{since } h \to 0)
$$
Thus:
$$
f'(x) = -6x - \frac{2}{x^2}
$$

### Step 5: Evaluate at $$ x = -1 $$
$$
f'(-1) = -6(-1) - \frac{2}{(-1)^2} = 6 - 2 = 4
$$

### Final Answer
The derivative of $$ f(x) $$ at $$ x = -1 $$ is **4**.
The derivative of $$ f(x) = -3x^{2} + \frac{2}{x} $$ at $$ x = -1 $$ is **4**.

(a) Using the first principles of differentiation, find the first derivative of \( f(x)=-3 x^{2}+\frac{2}{x} \) at \( x=-1 \).

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