find f(x). f(x)=\\sqrt3{(x^{2}+3)^{2}} f(x)=□

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\sqrt[3]{(x^{2}+3)^{2}}=(x^{2}+3)^{\frac{2}{3}}$ using the rule $\sqrt[n]{a^{m}} = a^{\frac{m}{n}}$.

Step2: Find the first - derivative using the chain - rule

The chain - rule states that if $y = u^{\frac{2}{3}}$ and $u=x^{2}+3$, then $y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{dy}{du}=\frac{2}{3}u^{-\frac{1}{3}}$ and $\frac{du}{dx} = 2x$. So, $f^\prime(x)=\frac{2}{3}(x^{2}+3)^{-\frac{1}{3}}\cdot2x=\frac{4x}{3(x^{2}+3)^{\frac{1}{3}}}$.

Step3: Find the second - derivative using the quotient - rule

The quotient - rule states that if $y=\frac{g(x)}{h(x)}$, then $y^\prime=\frac{g^\prime(x)h(x)-g(x)h^\prime(x)}{h(x)^{2}}$. Here, $g(x) = 4x$, $g^\prime(x)=4$, $h(x)=3(x^{2}+3)^{\frac{1}{3}}$, and $h^\prime(x)=3\times\frac{1}{3}(x^{2}+3)^{-\frac{2}{3}}\cdot2x = 2x(x^{2}+3)^{-\frac{2}{3}}$. [ \begin{align*} f^{\prime\prime}(x)&=\frac{4\times3(x^{2}+3)^{\frac{1}{3}}-4x\times2x(x^{2}+3)^{-\frac{2}{3}}}{9(x^{2}+3)^{\frac{2}{3}}}\ &=\frac{12(x^{2}+3)^{\frac{1}{3}}-\frac{8x^{2}}{(x^{2}+3)^{\frac{2}{3}}}}{9(x^{2}+3)^{\frac{2}{3}}}\ &=\frac{\frac{12(x^{2}+3)-8x^{2}}{(x^{2}+3)^{\frac{2}{3}}}}{9(x^{2}+3)^{\frac{2}{3}}}\ &=\frac{12x^{2}+36 - 8x^{2}}{9(x^{2}+3)^{\frac{4}{3}}}\ &=\frac{4x^{2}+36}{9(x^{2}+3)^{\frac{4}{3}}} \end{align*} ]

Answer:

$\frac{4x^{2}+36}{9(x^{2}+3)^{\frac{4}{3}}}$