5. $y = sqrt3{x}(x^{3}+1)^{5}$, determine $y$

Explanation:

Step1: Rewrite the function

Rewrite (\sqrt[3]{x}) as (x^{\frac{1}{3}}), so (y = x^{\frac{1}{3}}(x^{3}+1)^{5}).

Step2: Apply the product rule

The product rule is ((uv)^\prime=u^\prime v + uv^\prime). Let (u = x^{\frac{1}{3}}) and (v=(x^{3}+1)^{5}). First, find (u^\prime): (u^\prime=\frac{1}{3}x^{-\frac{2}{3}}) (using the power rule ((x^n)^\prime = nx^{n - 1})). Second, find (v^\prime) using the chain rule. Let (t=x^{3}+1), then (v = t^{5}). The chain rule (\frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}). (\frac{dv}{dt}=5t^{4}) and (\frac{dt}{dx}=3x^{2}), so (v^\prime = 5(x^{3}+1)^{4}\cdot3x^{2}=15x^{2}(x^{3}+1)^{4}).

Step3: Calculate (y^\prime)

(y^\prime=u^\prime v+uv^\prime=\frac{1}{3}x^{-\frac{2}{3}}(x^{3}+1)^{5}+x^{\frac{1}{3}}\cdot15x^{2}(x^{3}+1)^{4}). Simplify the first term: (\frac{1}{3}x^{-\frac{2}{3}}(x^{3}+1)^{5}=\frac{(x^{3}+1)^{5}}{3x^{\frac{2}{3}}}). Simplify the second term: (x^{\frac{1}{3}}\cdot15x^{2}(x^{3}+1)^{4}=15x^{\frac{7}{3}}(x^{3}+1)^{4}). Factor out (\frac{(x^{3}+1)^{4}}{3x^{\frac{2}{3}}}): (y^\prime=\frac{(x^{3}+1)^{4}}{3x^{\frac{2}{3}}}\left[(x^{3}+1)+45x^{3}\right]=\frac{(x^{3}+1)^{4}(46x^{3}+1)}{3x^{\frac{2}{3}}}).

Answer:

(y^\prime=\frac{(x^{3}+1)^{4}(46x^{3}+1)}{3x^{\frac{2}{3}}})