Find \( \frac{d y}{d x} \) \( y=\csc x \cot x \) \( \frac{d y}{d x}=\square \)

Given \[ y=\csc x\,\cot x, \] we differentiate \(y\) with respect to \(x\) using the product rule. Let \[ u=\csc x \quad \text{and} \quad v=\cot x. \] **Step 1.** Differentiate \(u\) and \(v\): \[ u'=\frac{d}{dx}(\csc x)=-\csc x\,\cot x, \] \[ v'=\frac{d}{dx}(\cot x)=-\csc^2 x. \] **Step 2.** Apply the product rule: \[ \frac{dy}{dx}=u'\,v+u\,v' = \left(-\csc x\,\cot x\right)(\cot x) + (\csc x)\left(-\csc^2 x\right). \] **Step 3.** Simplify the expression: \[ \frac{dy}{dx}=-\csc x\,\cot^2 x-\csc^3 x. \] Factor out \(\csc x\): \[ \frac{dy}{dx}=-\csc x\left(\cot^2 x+\csc^2 x\right). \] **Step 4.** Use the trigonometric identity: \[ \csc^2 x=1+\cot^2 x \quad \Longrightarrow \quad \cot^2 x=\csc^2 x-1. \] Substitute into the expression: \[ \cot^2 x+\csc^2 x=(\csc^2 x-1)+\csc^2 x=2\csc^2 x-1. \] **Final Answer:** \[ \frac{dy}{dx}=-\csc x\left(2\csc^2 x-1\right). \]