Question

let
$f(x)=(2x^{2}+3)^{3}(6x^{2}-3)^{9}$
$f(x)=$
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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 \(6x(2x^{2}+3)^{2}(6x^{2}-3)^{8}\):

$$
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$$

Answer:

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