let f(u)=∛u and g(x)=u = 2 + 6x². find (f ∘ g)(3). (f ∘ g)(3)=□ (type an exact answer.)

Explanation:

Step1: Find the derivative of $f(u)$

$f(u)=u^{\frac{1}{3}}$, so $f^\prime(u)=\frac{1}{3}u^{-\frac{2}{3}}$ by the power - rule $\frac{d}{du}(u^n)=nu^{n - 1}$.

Step2: Find the derivative of $g(x)$

$g(x)=2 + 6x^{2}$, so $g^\prime(x)=12x$ by the power - rule $\frac{d}{dx}(ax^n)=nax^{n - 1}$ and $\frac{d}{dx}(c)=0$ for a constant $c$.

Step3: Use the chain - rule

The chain - rule states that $(f\circ g)^\prime(x)=f^\prime(g(x))\cdot g^\prime(x)$. First, find $g(3)$: $g(3)=2+6\times3^{2}=2 + 54=56$. Then find $f^\prime(g(3))$: $f^\prime(g(3))=f^\prime(56)=\frac{1}{3}(56)^{-\frac{2}{3}}$. And $g^\prime(3)=12\times3 = 36$.

Step4: Calculate $(f\circ g)^\prime(3)$

$(f\circ g)^\prime(3)=f^\prime(g(3))\cdot g^\prime(3)=\frac{1}{3}(56)^{-\frac{2}{3}}\times36$. $(f\circ g)^\prime(3)= 12\times56^{-\frac{2}{3}}=12\times\frac{1}{(56^{\frac{1}{3}})^2}=12\times\frac{1}{(\sqrt[3]{56})^2}=12\times\frac{1}{(2\sqrt[3]{7})^2}=12\times\frac{1}{4\times7^{\frac{2}{3}}}=\frac{3}{7^{\frac{2}{3}}}=\frac{3}{\sqrt[3]{49}}$

Answer:

$\frac{3}{\sqrt[3]{49}}$