$$ y=a x^{2}+a $$ or determune $$ f^{\prime}(-1) $$ if $$ f(x)=5 x^{3}+2 x-\frac{1}{2 x^{3}} $$

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1. Write the function in a convenient form:
   $$
   f(x) = 5x^3 + 2x - \frac{1}{2x^3} = 5x^3 + 2x - \frac{1}{2} x^{-3}
   $$

2. Differentiate term by term:
   - The derivative of $$ 5x^3 $$ is:
     $$
     \frac{d}{dx}(5x^3) = 15x^2
     $$
   - The derivative of $$ 2x $$ is:
     $$
     \frac{d}{dx}(2x) = 2
     $$
   - The derivative of $$-\frac{1}{2}x^{-3}$$ is:
     $$
     \frac{d}{dx}\left( -\frac{1}{2}x^{-3} \right) = -\frac{1}{2} \cdot (-3)x^{-4} = \frac{3}{2}x^{-4}
     $$

3. Combine the results to get the derivative $$ f'(x) $$:
   $$
   f'(x) = 15x^2 + 2 + \frac{3}{2}x^{-4}
   $$

4. Evaluate the derivative at $$ x = -1 $$:
   - Compute each term:
     $$
     15(-1)^2 = 15 \cdot 1 = 15
     $$
     $$
     2 = 2
     $$
     $$
     \frac{3}{2}(-1)^{-4} = \frac{3}{2} \cdot 1 = \frac{3}{2}
     $$
   - Sum the terms:
     $$
     f'(-1) = 15 + 2 + \frac{3}{2} = 17 + \frac{3}{2} = \frac{34}{2} + \frac{3}{2} = \frac{37}{2}
     $$

Thus, the value of $$ f'(-1) $$ is $$ \frac{37}{2} $$.
$$ f'(-1) = \frac{37}{2} $$

or determune if

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