Question

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

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, $u = x^{3}$ and $v=\cos(x)$.

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

The derivative of $u = x^{3}$ with respect to $x$ is $\frac{du}{dx}=3x^{2}$ (using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$), and the derivative of $v=\cos(x)$ with respect to $x$ is $\frac{dv}{dx}=-\sin(x)$.

Step3: Substitute into quotient - rule

$\frac{dy}{dx}=\frac{\cos(x)\cdot3x^{2}-x^{3}\cdot(-\sin(x))}{\cos^{2}(x)}=\frac{3x^{2}\cos(x)+x^{3}\sin(x)}{\cos^{2}(x)}$

Answer:

$\frac{3x^{2}\cos(x)+x^{3}\sin(x)}{\cos^{2}(x)}$