find $\frac{dy}{dx}$. \n$y = 4(cot x+csc x)(cot x - csc x)$\n$\frac{dy}{dx}=square$

find $\frac{dy}{dx}$. \n$y = 4(cot x+csc x)(cot x - csc x)$\n$\frac{dy}{dx}=square$
Answer
Explanation:
Step1: Simplify the function using difference - of - squares
We know that $(a + b)(a - b)=a^{2}-b^{2}$. Here, $a = \cot x$ and $b=\csc x$, so $y = 4(\cot^{2}x-\csc^{2}x)$.
Step2: Use the trigonometric identity
Recall the identity $1+\cot^{2}x=\csc^{2}x$, then $\cot^{2}x-\csc^{2}x=- 1$. So $y=-4$.
Step3: Differentiate the constant function
The derivative of a constant function $y = C$ (where $C$ is a constant) with respect to $x$ is $0$. Since $y=-4$, $\frac{dy}{dx}=0$.
Answer:
$0$