Question

homework4: problem 13 (1 point) let $f(x)=\frac{1}{(x^{3}-\frac{3}{x})^{5}}$. find $f(x)$. $f(x)=square$

Explanation:

Step1: Rewrite the function

Rewrite $f(x)$ as $f(x)=(x^{3}-\frac{3}{x})^{- 5}$.

Step2: Apply the chain - rule

Let $u = x^{3}-3x^{-1}$, then $f(x)=u^{-5}$. The chain - rule states that $f^{\prime}(x)=\frac{df}{du}\cdot\frac{du}{dx}$. First, find $\frac{df}{du}$:
$\frac{df}{du}=-5u^{-6}$ (using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$ with $n=-5$).

Step3: Find $\frac{du}{dx}$

$\frac{du}{dx}=\frac{d}{dx}(x^{3}-3x^{-1}) = 3x^{2}+3x^{-2}$ (using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ for $n = 3$ and $n=-1$).

Step4: Calculate $f^{\prime}(x)$

$f^{\prime}(x)=\frac{df}{du}\cdot\frac{du}{dx}=-5u^{-6}\cdot(3x^{2}+3x^{-2})$. Substitute $u = x^{3}-3x^{-1}$ back in:
$f^{\prime}(x)=-5(x^{3}-\frac{3}{x})^{-6}\cdot(3x^{2}+\frac{3}{x^{2}})=-\frac{5(3x^{2}+\frac{3}{x^{2}})}{(x^{3}-\frac{3}{x})^{6}}$.

Answer:

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