question 13 · 1 point\nfind the derivative of $f(x)=4x^{-2}cdotln(x^{4}-3x)$.\nprovide your answer…

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = 4x^{-2}$ and $v=\ln(x^{4}-3x)$. First, find $u^\prime$ and $v^\prime$. $u^\prime=\frac{d}{dx}(4x^{-2})=-8x^{-3}$

Step2: Apply chain - rule for $v^\prime$

The chain - rule states that if $y = \ln(u)$ and $u = x^{4}-3x$, then $\frac{dy}{dx}=\frac{1}{u}\cdot\frac{du}{dx}$. $\frac{du}{dx}=\frac{d}{dx}(x^{4}-3x)=4x^{3}-3$, so $v^\prime=\frac{4x^{3}-3}{x^{4}-3x}$

Step3: Calculate $f^\prime(x)$

Using the product - rule $f^\prime(x)=u^\prime v+uv^\prime$. $f^\prime(x)=-8x^{-3}\ln(x^{4}-3x)+4x^{-2}\cdot\frac{4x^{3}-3}{x^{4}-3x}$ $=-\frac{8\ln(x^{4}-3x)}{x^{3}}+\frac{4(4x^{3}-3)}{x^{2}(x^{4}-3x)}$

Answer:

$-\frac{8\ln(x^{4}-3x)}{x^{3}}+\frac{4(4x^{3}-3)}{x^{2}(x^{4}-3x)}$