Question

differentiate the following function.
f(x)=3c^{5}(x^{2}+3)^{2}
f(x)=\square

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 \(u = 3e^{5}\) (a constant, so \(u^\prime=0\)) and \(v=(x^{2}+3)^{2}\).

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

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 \(u^\prime = 0\) and \(v=(x^{2}+3)^{2}\), \(u = 3e^{5}\), \(v^\prime=4x(x^{2}+3)\)

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Answer:

\(12e^{5}(x^{3}+3x)\)