find ( f^{prime prime}(x) ).\n\n( f(x)=left(x^{2}+7\right)^{9} )\n\n( 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))