Question

find $\frac{dy}{dx}$. $y = 3cos xcot x$ $\frac{dy}{dx}=square$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $x$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. Here, $u = 3\cos x$ and $v=\cot x$.

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

We know that $\frac{d}{dx}(\cos x)=-\sin x$, so $\frac{d}{dx}(3\cos x)=- 3\sin x$. Also, $\frac{d}{dx}(\cot x)=-\csc^{2}x$.

Step3: Substitute into product - rule

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

Step4: Simplify the expression

$3\cos x(-\csc^{2}x)+\cot x(-3\sin x)=-3\frac{\cos x}{\sin^{2}x}-3\frac{\cos x}{\sin x}\cdot\sin x=-3\frac{\cos x}{\sin^{2}x}-3\cos x=-3\cos x(\frac{1 + \sin^{2}x}{\sin^{2}x})$.

Answer:

$-3\cos x\csc^{2}x - 3\cos x$