Question

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

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:

$0$