Find \( f^{\prime \prime}(x) \) \( f(x)=\left(x^{2}+4\right)^{6} \) \( f^{\prime \prime}(x)=\square \)
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To find \( f^{\prime \prime}(x) \) for the function \( f(x) = (x^2 + 4)^6 \), we first need to calculate the first derivative \( f'(x) \) using the chain rule. 1. **First Derivative**: \[ f'(x) = \frac{d}{dx}\left((x^2 + 4)^6\right) = 6(x^2 + 4)^5 \cdot \frac{d}{dx}(x^2 + 4) = 6(x^2 + 4)^5 \cdot 2x = 12x(x^2 + 4)^5. \] 2. **Second Derivative**: Now we apply the product rule and the chain rule again to differentiate \( f'(x) \): \[ f^{\prime\prime}(x) = \frac{d}{dx}[12x(x^2 + 4)^5]. \] Let \( u = 12x \) and \( v = (x^2 + 4)^5 \). Thus, we compute: \[ f^{\prime\prime}(x) = u'v + uv'. \] - First, differentiate \( u \): \[ u' = 12. \] - Second, differentiate \( v \) using the chain rule: \[ v' = 5(x^2 + 4)^4 \cdot \frac{d}{dx}(x^2 + 4) = 5(x^2 + 4)^4 \cdot 2x = 10x(x^2 + 4)^4. \] Now, substitute into the product rule: \[ f^{\prime\prime}(x) = 12(x^2 + 4)^5 + 12x \cdot 10x(x^2 + 4)^4. \] Simplifying this gives: \[ f^{\prime\prime}(x) = 12(x^2 + 4)^5 + 120x^2(x^2 + 4)^4. \] Factoring out common terms: \[ f^{\prime\prime}(x) = 12(x^2 + 4)^4\left((x^2 + 4) + 10x^2\right) = 12(x^2 + 4)^4(11x^2 + 4). \] Thus, the second derivative is: \[ f^{\prime\prime}(x) = 12(x^2 + 4)^4(11x^2 + 4). \]
