test: mid - term math 1325 spring 25 differentiate the function. g(x)=\\sqrt4{x^{3}-6x} g(x)=□

Explanation:

Step1: Rewrite the function

Rewrite $G(x)=\sqrt[4]{x^{3}-6x}=(x^{3}-6x)^{\frac{1}{4}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^{\frac{1}{4}}$ and $u=x^{3}-6x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$: $\frac{d}{du}(u^{\frac{1}{4}})=\frac{1}{4}u^{-\frac{3}{4}}$. Second, find $\frac{du}{dx}$: $\frac{d}{dx}(x^{3}-6x)=3x^{2}-6$.

Step3: Substitute $u$ back

Substitute $u = x^{3}-6x$ into $\frac{dy}{du}\cdot\frac{du}{dx}$. We get $G^{\prime}(x)=\frac{1}{4}(x^{3}-6x)^{-\frac{3}{4}}\cdot(3x^{2}-6)$.

Step4: Simplify the expression

$G^{\prime}(x)=\frac{3x^{2}-6}{4\sqrt[4]{(x^{3}-6x)^{3}}}$.

Answer:

$\frac{3x^{2}-6}{4\sqrt[4]{(x^{3}-6x)^{3}}}$