Question

given the function $y =-\frac{2sqrt3{x^{2}}}{3}$, find $\frac{dy}{dx}$. express your answer in radical form without using negative exponents, simplifying all fractions.

Explanation:

Step1: Rewrite the function

We can rewrite $y =-\frac{2\sqrt[3]{x^{2}}}{3}=-\frac{2}{3}x^{\frac{2}{3}}$.

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = ax^{n}$, then $\frac{dy}{dx}=anx^{n - 1}$. Here, $a =-\frac{2}{3}$ and $n=\frac{2}{3}$. So, $\frac{dy}{dx}=-\frac{2}{3}\times\frac{2}{3}x^{\frac{2}{3}-1}$.

Step3: Simplify the exponent and the coefficient

First, simplify the coefficient: $-\frac{2}{3}\times\frac{2}{3}=-\frac{4}{9}$. Then, simplify the exponent: $\frac{2}{3}-1=\frac{2 - 3}{3}=-\frac{1}{3}$. So, $\frac{dy}{dx}=-\frac{4}{9}x^{-\frac{1}{3}}$.

Step4: Rewrite without negative exponents

Using the rule $x^{-n}=\frac{1}{x^{n}}$, we get $\frac{dy}{dx}=-\frac{4}{9x^{\frac{1}{3}}}=-\frac{4}{9\sqrt[3]{x}}$.

Answer:

$-\frac{4}{9\sqrt[3]{x}}$