differentiate the following function.\n\n$f(x)=4c^{9}(x^{2}+3)^{2}$\n\n$f(x)=\\square$

Explanation:

Step1: Apply the product rule

The product rule states that if (y = uv), then (y^\prime=u^\prime v + uv^\prime). Let (u = 4c^{9}) (a constant) and (v=(x^{2}+3)^{2}). Since (u^\prime = 0) (derivative of a constant is (0)), we only need to find (v^\prime) using the chain rule.

Step2: Apply the chain rule to (v=(x^{2}+3)^{2})

The chain rule: if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)). Let (g(x)=x^{2}+3), (f(u) = u^{2}). (f^\prime(u) = 2u), (g^\prime(x)=2x). So (v^\prime=2(x^{2}+3)\cdot2x = 4x(x^{2}+3))

Step3: Calculate (f^\prime(x))

Since (f(x)=uv) and (u^\prime = 0), (f^\prime(x)=u^\prime v+uv^\prime=0\times(x^{2}+3)^{2}+4c^{9}\times4x(x^{2}+3))

Answer:

(16c^{9}x(x^{2}+3))