Question

differentiate the following function.

$f(x)=4c^{9}(x^{2}+3)^{2}$

$f(x)=\square$

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:

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