Question

differentiate the function g(x)=\sqrt5{x^{3}-4x} g(x)=□

Explanation:

Step1: Rewrite the function

Rewrite $G(x)=\sqrt[5]{x^{3}-4x}=(x^{3}-4x)^{\frac{1}{5}}$ using the rule $\sqrt[n]{a}=a^{\frac{1}{n}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = f(u)$ and $u = g(x)$, then $y^\prime=f^\prime(u)\cdot g^\prime(x)$. Let $u = x^{3}-4x$ and $y = u^{\frac{1}{5}}$. First, find the derivative of $y$ with respect to $u$: $y^\prime_{u}=\frac{1}{5}u^{-\frac{4}{5}}$ (using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$). Then find the derivative of $u$ with respect to $x$: $u^\prime_{x}=3x^{2}-4$ (using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ for $x^{3}$ and $\frac{d}{dx}(4x)=4$).

Step3: Multiply the derivatives

By the chain - rule, $G^\prime(x)=y^\prime_{u}\cdot u^\prime_{x}=\frac{1}{5}(x^{3}-4x)^{-\frac{4}{5}}\cdot(3x^{2}-4)$.

Step4: Simplify the result

$G^\prime(x)=\frac{3x^{2}-4}{5\sqrt[5]{(x^{3}-4x)^{4}}}$

Answer:

$\frac{3x^{2}-4}{5\sqrt[5]{(x^{3}-4x)^{4}}}$