find f(x). f(x)=\\sqrt7{(x^{2}+9)^{6}} f(x)=□

Explanation:

Step1: Rewrite the function

First, rewrite $f(x)=\sqrt[7]{(x^{2}+9)^{6}}=(x^{2}+9)^{\frac{6}{7}}$ using the radical - exponent rule $\sqrt[n]{a^{m}}=a^{\frac{m}{n}}$.

Step2: Find the first - derivative using the chain rule

The chain rule states that if $y = u^{\frac{6}{7}}$ and $u=x^{2}+9$, then $y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{dy}{du}=\frac{6}{7}u^{-\frac{1}{7}}$ and $\frac{du}{dx} = 2x$. So, $f^\prime(x)=\frac{6}{7}(x^{2}+9)^{-\frac{1}{7}}\cdot2x=\frac{12x}{7(x^{2}+9)^{\frac{1}{7}}}$.

Step3: Find the second - derivative using the quotient rule

The quotient rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u = 12x$, $u^\prime=12$, $v = 7(x^{2}+9)^{\frac{1}{7}}$, and $v^\prime=7\times\frac{1}{7}(x^{2}+9)^{-\frac{6}{7}}\cdot2x=\frac{2x}{(x^{2}+9)^{\frac{6}{7}}}$. [ \begin{align*} f^{\prime\prime}(x)&=\frac{12\times7(x^{2}+9)^{\frac{1}{7}}-12x\times\frac{2x}{(x^{2}+9)^{\frac{6}{7}}}\times7}{49(x^{2}+9)^{\frac{2}{7}}}\ &=\frac{84(x^{2}+9)^{\frac{1}{7}}-\frac{168x^{2}}{(x^{2}+9)^{\frac{6}{7}}}}{49(x^{2}+9)^{\frac{2}{7}}}\ &=\frac{84(x^{2}+9)-168x^{2}}{49(x^{2}+9)^{\frac{2}{7}+\frac{6}{7}}}\ &=\frac{84x^{2}+756 - 168x^{2}}{49(x^{2}+9)^{\frac{8}{7}}}\ &=\frac{756 - 84x^{2}}{49(x^{2}+9)^{\frac{8}{7}}} \end{align*} ]

Answer:

$\frac{756 - 84x^{2}}{49(x^{2}+9)^{\frac{8}{7}}}$