Question

given the function $y = 3x^{-3}cos x$, find $\frac{dy}{dx}$ in any form.

Explanation:

Step1: Apply product - rule

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

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

Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, for $u = 3x^{-3}$, we have $\frac{du}{dx}=3\times(-3)x^{-3 - 1}=-9x^{-4}$.

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

The derivative of $\cos x$ with respect to $x$ is $-\sin x$, so $\frac{dv}{dx}=-\sin x$.

Step4: Substitute into product - rule

$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}=3x^{-3}(-\sin x)+\cos x(-9x^{-4})$.

Step5: Simplify

$\frac{dy}{dx}=- 3x^{-3}\sin x-9x^{-4}\cos x=-\frac{3\sin x}{x^{3}}-\frac{9\cos x}{x^{4}}$.

Answer:

$-\frac{3\sin x}{x^{3}}-\frac{9\cos x}{x^{4}}$