Question

find ( f^{prime prime}(x) ).

( f(x)=left(x^{2}+7
ight)^{9} )

( f^{prime prime}(x)= )

Explanation:

Step1: Find the first derivative using the chain rule

Let \(u = x^{2}+7\), then \(y = u^{9}\).
The chain rule states \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\).
\(\frac{dy}{du}=9u^{8}\) and \(\frac{du}{dx}=2x\).
So \(f^{\prime}(x)=9(x^{2}+7)^{8}\cdot2x = 18x(x^{2}+7)^{8}\).

Step2: Find the second derivative using the product rule

The product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 18x\) and \(v=(x^{2}+7)^{8}\).
\(u^\prime=18\), and for \(v^\prime\), use the chain rule again. Let \(t=x^{2}+7\), \(v = t^{8}\), \(\frac{dv}{dt}=8t^{7}\), \(\frac{dt}{dx}=2x\), so \(v^\prime=8(x^{2}+7)^{7}\cdot2x = 16x(x^{2}+7)^{7}\).
\(f^{\prime\prime}(x)=18(x^{2}+7)^{8}+18x\cdot16x(x^{2}+7)^{7}\)
\(=18(x^{2}+7)^{8}+288x^{2}(x^{2}+7)^{7}\)
Factor out \(18(x^{2}+7)^{7}\):
\(f^{\prime\prime}(x)=18(x^{2}+7)^{7}[(x^{2}+7)+16x^{2}]\)
\(=18(x^{2}+7)^{7}(17x^{2}+7)\)

Answer:

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