Question

let $y = \frac{\cos(x)}{x^{3}}$.
$\frac{dy}{dx}=$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$. Here, $u = \cos(x)$ and $v=x^{3}$.

Step2: Find $\frac{du}{dx}$ and $\frac{dv}{dx}$

The derivative of $\cos(x)$ with respect to $x$ is $-\sin(x)$, so $\frac{du}{dx}=-\sin(x)$. The derivative of $x^{3}$ with respect to $x$ is $3x^{2}$, so $\frac{dv}{dx}=3x^{2}$.

Step3: Substitute into quotient - rule

$\frac{dy}{dx}=\frac{x^{3}(-\sin(x))-\cos(x)\times3x^{2}}{(x^{3})^{2}}$.

Step4: Simplify the expression

$\frac{dy}{dx}=\frac{-x^{3}\sin(x)- 3x^{2}\cos(x)}{x^{6}}=\frac{-x\sin(x)-3\cos(x)}{x^{4}}$.

Answer:

$\frac{-x\sin(x)-3\cos(x)}{x^{4}}$