given the function $y = \frac{4}{sqrt3{x^{3}}}$, find $\frac{dy}{dx}$. express your answer in radical form…

Explanation:

Step1: Rewrite the function

Rewrite $y = \frac{4}{\sqrt[3]{x^{5}}}$ as $y = 4x^{-\frac{5}{3}}$ using the rule $\frac{1}{x^{n}}=x^{-n}$ and $\sqrt[m]{x^{n}}=x^{\frac{n}{m}}$.

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = ax^{n}$, then $\frac{dy}{dx}=anx^{n - 1}$. Here, $a = 4$ and $n=-\frac{5}{3}$. So, $\frac{dy}{dx}=4\times(-\frac{5}{3})x^{-\frac{5}{3}-1}$.

Step3: Simplify the exponent and coefficient

First, calculate the exponent: $-\frac{5}{3}-1=-\frac{5}{3}-\frac{3}{3}=-\frac{8}{3}$. Then, the derivative is $\frac{dy}{dx}=-\frac{20}{3}x^{-\frac{8}{3}}$.

Step4: Rewrite without negative exponents

Using the rule $x^{-n}=\frac{1}{x^{n}}$, we get $\frac{dy}{dx}=-\frac{20}{3x^{\frac{8}{3}}}$. And $x^{\frac{8}{3}}=\sqrt[3]{x^{8}}$, so $\frac{dy}{dx}=-\frac{20}{3\sqrt[3]{x^{8}}}$.

Answer:

$-\frac{20}{3\sqrt[3]{x^{8}}}$