Find $$ f^{\prime \prime}(x) $$ $$ f(x)=\left(x^{2}+4\right)^{6} $$ $$ f^{\prime \prime}(x)=\square $$

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1. We start with
   $$
   f(x)=\left(x^2+4\right)^6.
   $$

2. Compute the first derivative $$ f'(x) $$ using the chain rule:
   $$
   f'(x)=6\left(x^2+4\right)^5\cdot (2x)=12x\left(x^2+4\right)^5.
   $$

3. To find the second derivative $$ f''(x) $$, differentiate $$ f'(x) $$ using the product rule. Write
   $$
   f'(x)=12x\cdot\left(x^2+4\right)^5.
   $$
   The product rule gives us:
   $$
   f''(x)=12\left(x^2+4\right)^5 + 12x\cdot\frac{d}{dx}\left[\left(x^2+4\right)^5\right].
   $$

4. Differentiate $$\left(x^2+4\right)^5$$ using the chain rule:
   $$
   \frac{d}{dx}\left[\left(x^2+4\right)^5\right]=5\left(x^2+4\right)^4\cdot (2x)=10x\left(x^2+4\right)^4.
   $$

5. Substitute back into the expression for $$ f''(x) $$:
   $$
   f''(x)=12\left(x^2+4\right)^5 + 12x\cdot \left(10x\left(x^2+4\right)^4\right).
   $$
   Simplify the second term:
   $$
   12x\cdot 10x\left(x^2+4\right)^4=120x^2\left(x^2+4\right)^4.
   $$

6. Thus, 
   $$
   f''(x)=12\left(x^2+4\right)^5 + 120x^2\left(x^2+4\right)^4.
   $$

7. Factor out the common factor $$ 12\left(x^2+4\right)^4 $$:
   $$
   f''(x)=12\left(x^2+4\right)^4\left[\left(x^2+4\right) + 10x^2\right].
   $$

8. Combine like terms inside the bracket:
   $$
   \left(x^2+4\right)+10x^2=11x^2+4.
   $$

9. The final expression for the second derivative is:
   $$
   f''(x)=12\left(x^2+4\right)^4\left(11x^2+4\right).
   $$
$$ f''(x) = 12(x^2 + 4)^4 (11x^2 + 4) $$

Find \( f^{\prime \prime}(x) \) \( f(x)=\left(x^{2}+4\right)^{6} \) \( f^{\prime \prime}(x)=\square \)

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