Question

given the function $f(x)=\frac{1}{sqrt3{6x^{2}+4x + 5}}$, find $f(x)$ in any form.

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\frac{1}{\sqrt[3]{6x^{2}+4x + 5}}$ as $f(x)=(6x^{2}+4x + 5)^{-\frac{1}{3}}$ using the negative - exponent rule $\frac{1}{a}=a^{-1}$ and the radical - exponent rule $\sqrt[n]{a}=a^{\frac{1}{n}}$.

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 $u = 6x^{2}+4x + 5$ and $n=-\frac{1}{3}$. First, find the derivative of $u$ with respect to $x$: $u^\prime=\frac{d}{dx}(6x^{2}+4x + 5)=12x + 4$. 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 $f^\prime(x)$

By the chain rule $f^\prime(x)=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $u = 6x^{2}+4x + 5$ and its derivatives: $f^\prime(x)=-\frac{1}{3}(6x^{2}+4x + 5)^{-\frac{4}{3}}\cdot(12x + 4)=-\frac{12x + 4}{3(6x^{2}+4x + 5)^{\frac{4}{3}}}$.

Answer:

$-\frac{12x + 4}{3(6x^{2}+4x + 5)^{\frac{4}{3}}}$