let\n$f(x)=(2x^{2}+3)^{3}(6x^{2}-3)^{9}$\n$f(x)=$\nquestion help: video message instruc\nsubmit question…

Answer:

$6x(2x^{2}+3)^{2}(6x^{2}-3)^{8}(22x^{2}-3)$

Explanation:

Step1: Apply the product rule

The product rule states that if (y = u\cdot v), then (y^\prime=u^\prime v + uv^\prime). Let (u=(2x^{2}+3)^{3}) and (v=(6x^{2}-3)^{9}).

Step2: Apply the chain rule to find (u^\prime)

The chain rule states that if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)). For (u=(2x^{2}+3)^{3}), let (g(x)=2x^{2}+3), (f(g)=g^{3}). Then (g^\prime(x) = 4x) and (f^\prime(g)=3g^{2}). So (u^\prime=3(2x^{2}+3)^{2}\cdot4x = 12x(2x^{2}+3)^{2})

Step3: Apply the chain rule to find (v^\prime)

For (v=(6x^{2}-3)^{9}), let (g(x)=6x^{2}-3), (f(g)=g^{9}). Then (g^\prime(x)=12x) and (f^\prime(g) = 9g^{8}). So (v^\prime=9(6x^{2}-3)^{8}\cdot12x=108x(6x^{2}-3)^{8})

Step4: Substitute (u^\prime), (u), (v^\prime), (v) into the product rule

(f^\prime(x)=u^\prime v+uv^\prime=12x(2x^{2}+3)^{2}(6x^{2}-3)^{9}+(2x^{2}+3)^{3}\cdot108x(6x^{2}-3)^{8})

Step5: Factor out common terms

Factor out (6x(2x^{2}+3)^{2}(6x^{2}-3)^{8}):

[ \begin{align*} f^\prime(x)&=6x(2x^{2}+3)^{2}(6x^{2}-3)^{8}[2(6x^{2}-3)+ 18(2x^{2}+3)]\ &=6x(2x^{2}+3)^{2}(6x^{2}-3)^{8}(12x^{2}-6 + 36x^{2}+54)\ &=6x(2x^{2}+3)^{2}(6x^{2}-3)^{8}(48x^{2}+48 - 6x^{2})\ &=6x(2x^{2}+3)^{2}(6x^{2}-3)^{8}(22x^{2}-3) \end{align*} ]