y = \\sqrt5{8x^{3}+9x}, (3,3)\ny(3) = \n

Explanation:

Step1: Rewrite the function

Rewrite $y = \sqrt[5]{8x^{3}+9x}$ as $y=(8x^{3}+9x)^{\frac{1}{5}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{5}}$ and $u = 8x^{3}+9x$, then $y^\prime=\frac{1}{5}u^{-\frac{4}{5}}\cdot u^\prime$. First, find $u^\prime$: $u^\prime=\frac{d}{dx}(8x^{3}+9x)=24x^{2}+9$. Then $y^\prime=\frac{1}{5}(8x^{3}+9x)^{-\frac{4}{5}}\cdot(24x^{2}+9)$.

Step3: Evaluate $y^\prime$ at $x = 3$

Substitute $x = 3$ into $y^\prime$. First, when $x = 3$, $8x^{3}+9x=8\times3^{3}+9\times3=8\times27 + 27=216+27 = 243$, and $24x^{2}+9=24\times3^{2}+9=24\times9 + 9=216+9=225$. Then $y^\prime(3)=\frac{1}{5}(243)^{-\frac{4}{5}}\times225$. Since $243 = 3^{5}$, $(243)^{-\frac{4}{5}}=(3^{5})^{-\frac{4}{5}}=3^{-4}=\frac{1}{81}$. So $y^\prime(3)=\frac{1}{5}\times\frac{1}{81}\times225=\frac{225}{405}=\frac{5}{9}$.

Answer:

$\frac{5}{9}$