find f(x).\nf(x)=4x^{3}ln^{2}x\nf(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 = 4x^{3}) and (v=\ln^{2}x). First, find (u^\prime): (u^\prime=\frac{d}{dx}(4x^{3}) = 12x^{2})

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

Let (t=\ln x), so (v = t^{2}). By the chain rule (\frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}). (\frac{dv}{dt} = 2t) and (\frac{dt}{dx}=\frac{1}{x}). Then (v^\prime=\frac{d}{dx}(\ln^{2}x)=2\ln x\cdot\frac{1}{x})

Step3: Substitute (u), (u^\prime), (v), (v^\prime) into the product rule formula

(f^\prime(x)=u^\prime v+uv^\prime) (f^\prime(x)=12x^{2}\cdot\ln^{2}x+4x^{3}\cdot\frac{2\ln x}{x})

Step4: Simplify the expression

(f^\prime(x)=12x^{2}\ln^{2}x + 8x^{2}\ln x) (f^\prime(x)=4x^{2}\ln x(3\ln x + 2))

Answer:

(4x^{2}\ln x(3\ln x + 2))