5. - / 1 points find the derivative of the function. f(x) = 1 / (∛(2x² - 3)) f(x) = - 4x /

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\frac{1}{\sqrt[3]{2x^{2}-3}}=(2x^{2}-3)^{-\frac{1}{3}}$ using negative - exponent rule.

Step2: Apply the chain - rule

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

Step3: Calculate the derivative of $f(x)$

By the chain - rule $f^\prime(x)=\frac{dy}{du}\cdot\frac{du}{dx}=-\frac{1}{3}(2x^{2}-3)^{-\frac{4}{3}}\cdot4x=-\frac{4x}{3(2x^{2}-3)^{\frac{4}{3}}}$.

Answer:

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