$$ f^{\prime}(x) $$ أوجدي $$ \cdot f(x)=2\left(\frac{2 x^{2}}{3}-3\right) $$

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Alright, I need to find the derivative $$ f^{\prime}(x) $$ of the function $$ f(x) = 2\left(\frac{2x^{2}}{3} - 3\right) $$. Let's break this down step by step to ensure accuracy.

First, I'll start by simplifying the expression inside the parentheses to make differentiation easier. The function is:

$$
f(x) = 2\left(\frac{2x^{2}}{3} - 3\right)
$$

I can distribute the 2 across the terms inside the parentheses:

$$
f(x) = 2 \times \frac{2x^{2}}{3} - 2 \times 3
$$

Calculating each term:

$$
2 \times \frac{2x^{2}}{3} = \frac{4x^{2}}{3}
$$
$$
2 \times 3 = 6
$$

So, the function simplifies to:

$$
f(x) = \frac{4x^{2}}{3} - 6
$$

Now, I need to find the derivative $$ f^{\prime}(x) $$. The derivative of a function measures the rate at which the function's value changes with respect to its variable, in this case, $$ x $$.

Let's differentiate each term separately.

1. **Differentiating $$ \frac{4x^{2}}{3} $$:**

The power rule of differentiation states that:

$$
\frac{d}{dx}x^{n} = n x^{n-1}
$$

Applying this to $$ \frac{4x^{2}}{3} $$:

$$
\frac{d}{dx}\left(\frac{4x^{2}}{3}\right) = \frac{4}{3} \times 2x^{1} = \frac{8x}{3}
$$

2. **Differentiating $$ -6 $$:**

The derivative of a constant is zero because constants do not change as $$ x $$ changes.

$$
\frac{d}{dx}(-6) = 0
$$

Combining both derivatives:

$$
f^{\prime}(x) = \frac{8x}{3} + 0 = \frac{8x}{3}
$$

So, the derivative of the function $$ f(x) = 2\left(\frac{2x^{2}}{3} - 3\right) $$ is:

$$
f^{\prime}(x) = \frac{8x}{3}
$$

To ensure there are no mistakes, I'll double-check the calculations:

- Simplifying the original function was correct.
- Applying the power rule to $$ \frac{4x^{2}}{3} $$ resulted in $$ \frac{8x}{3} $$, which seems accurate.
- The derivative of the constant term $$ -6 $$ is indeed zero.

Everything checks out, so the final derivative is confidently $$ \frac{8x}{3} $$.

$$ f^{\prime}(x) = \frac{8x}{3} $$

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