Given $$ f(x)=\frac{1}{\left(4-x^{2}\right)^{7}} $$, find $$ f^{\prime}(x) $$ $$ f^{\prime}(x)= $$

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To find the derivative $$ f'(x) $$ of the function $$ f(x) = \frac{1}{(4 - x^2)^7} $$, we can use the chain rule.

1. **Rewrite the function:**  
   $$ f(x) = (4 - x^2)^{-7} $$

2. **Differentiate using the chain rule:**  
   $$
   f'(x) = -7 \cdot (4 - x^2)^{-8} \cdot (-2x)
   $$
   Here, 
   - The outer function is $$ u^{-7} $$ with derivative $$ -7u^{-8} $$.
   - The inner function is $$ u = 4 - x^2 $$ with derivative $$ -2x $$.

3. **Simplify the expression:**  
   $$
   f'(x) = 14x \cdot (4 - x^2)^{-8}
   $$
   Or equivalently,
   $$
   f'(x) = \frac{14x}{(4 - x^2)^8}
   $$

**Final Answer:**
$$
f^{\prime}(x) = \frac{14\,x}{\,\left(4 - x^{2}\right)^{8}}
$$
$$ f^{\prime}(x) = \frac{14x}{(4 - x^2)^8} $$

Given \( f(x)=\frac{1}{\left(4-x^{2}\right)^{7}} \), find \( f^{\prime}(x) \) \( f^{\prime}(x)= \)

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