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