Question

given the function $y = \frac{4}{sqrt3{3x^{2}-3x}}$, find $\frac{dy}{dx}$ in any form.

Explanation:

Step1: Rewrite the function

We can rewrite $y=\frac{4}{\sqrt[3]{3x^{2}-3x}}$ as $y = 4(3x^{2}-3x)^{-\frac{1}{3}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. Let $u = 3x^{2}-3x$, so $y = 4u^{-\frac{1}{3}}$. First, find the derivative of $y$ with respect to $u$: $\frac{dy}{du}=4\times(-\frac{1}{3})u^{-\frac{1}{3}-1}=-\frac{4}{3}u^{-\frac{4}{3}}$. Then find the derivative of $u$ with respect to $x$: $\frac{du}{dx}=(3x^{2}-3x)^\prime = 6x - 3$.

Step3: Calculate $\frac{dy}{dx}$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}$ and $\frac{du}{dx}$: $\frac{dy}{dx}=-\frac{4}{3}(3x^{2}-3x)^{-\frac{4}{3}}\cdot(6x - 3)$. Simplify the expression: $\frac{dy}{dx}=-\frac{4(6x - 3)}{3(3x^{2}-3x)^{\frac{4}{3}}}=-\frac{4\times3(2x - 1)}{3(3x^{2}-3x)^{\frac{4}{3}}}=-\frac{4(2x - 1)}{(3x^{2}-3x)^{\frac{4}{3}}}$.

Answer:

$-\frac{4(2x - 1)}{(3x^{2}-3x)^{\frac{4}{3}}}$