Question

determine all critical

its for the following function.

f(x)=2x(16 - x)^3

what is the derivative of f(x)=2x(16 - x)^3?

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 \(u = 2x\) and \(v=(16 - x)^{3}\).
First, find \(u^\prime\): \(u^\prime=\frac{d}{dx}(2x)=2\).
Next, find \(v^\prime\) using the chain rule. If \(y=(16 - x)^{3}\), let \(t = 16 - x\), then \(y = t^{3}\). \(\frac{dy}{dt}=3t^{2}\) and \(\frac{dt}{dx}=- 1\). So \(v^\prime=\frac{d}{dx}(16 - x)^{3}=3(16 - x)^{2}\cdot(-1)=-3(16 - x)^{2}\).

Step2: Substitute into the product rule formula

\(f^\prime(x)=u^\prime v+uv^\prime\)
\(f^\prime(x)=2(16 - x)^{3}+2x\cdot(-3)(16 - x)^{2}\)

Step3: Factor out the common term

Factor out \(2(16 - x)^{2}\):
\(f^\prime(x)=2(16 - x)^{2}[(16 - x)-3x]\)
Simplify the expression inside the brackets: \((16 - x)-3x=16 - 4x = 4(4 - x)\)
So \(f^\prime(x)=8(16 - x)^{2}(4 - x)\)

Answer:

\(8(16 - x)^{2}(4 - x)\)