Question

find f(x).
f(x) = (x^2 + 8)^9

Explanation:

Step1: Find the first - derivative using the chain - rule

The chain - rule states that if \(y = u^n\) and \(u = g(x)\), then \(y^\prime=n\cdot u^{n - 1}\cdot g^\prime(x)\). Let \(u=x^{2}+8\) and \(n = 9\). Then \(f^\prime(x)=9(x^{2}+8)^{8}\cdot(2x)=18x(x^{2}+8)^{8}\).

Step2: Find the second - derivative using the product - rule

The product - rule states that if \(y = uv\), where \(u\) and \(v\) are functions of \(x\), then \(y^\prime=u^\prime v+uv^\prime\). Let \(u = 18x\) and \(v=(x^{2}+8)^{8}\).
First, find \(u^\prime\) and \(v^\prime\). \(u^\prime = 18\). To find \(v^\prime\), use the chain - rule again. Let \(t=x^{2}+8\), so \(v = t^{8}\). Then \(v^\prime=8t^{7}\cdot(2x)=16x(x^{2}+8)^{7}\).
Now, apply the product - rule:
\[

$$\begin{align*} f^{\prime\prime}(x)&=u^\prime v+uv^\prime\\ &=18(x^{2}+8)^{8}+18x\cdot16x(x^{2}+8)^{7}\\ &=18(x^{2}+8)^{7}[(x^{2}+8)+16x^{2}]\\ &=18(x^{2}+8)^{7}(17x^{2}+8) \end{align*}$$

\]

Answer:

\(18(x^{2}+8)^{7}(17x^{2}+8)\)