Show that if \( y=\operatorname{cosec} x \) then \( \frac{d y}{d x}=-\operatorname{cosec} x \cot x \)

To show that if \( y = \operatorname{cosec} x \), then \( \frac{dy}{dx} = -\operatorname{cosec} x \cot x \), we will differentiate \( y \) with respect to \( x \). Recall that the cosecant function is defined as: \[ \operatorname{cosec} x = \frac{1}{\sin x} \] Now, we can differentiate \( y \) using the quotient rule or the chain rule. Here, we will use the chain rule. First, we rewrite \( y \): \[ y = \sin x^{-1} \] Now, we apply the chain rule: \[ \frac{dy}{dx} = -\sin^{-2} x \cdot \frac{d}{dx}(\sin x) \] The derivative of \( \sin x \) is \( \cos x \), so we have: \[ \frac{dy}{dx} = -\sin^{-2} x \cdot \cos x \] Now, we can rewrite \( \sin^{-2} x \) as \( \operatorname{cosec}^2 x \): \[ \frac{dy}{dx} = -\operatorname{cosec}^2 x \cdot \cos x \] Next, we can express \( \cos x \) in terms of \( \operatorname{cosec} x \) and \( \cot x \): \[ \cos x = \sin x \cot x \] Substituting this back into our derivative gives: \[ \frac{dy}{dx} = -\operatorname{cosec}^2 x \cdot \sin x \cot x \] Since \( \operatorname{cosec} x = \frac{1}{\sin x} \), we can express \( \operatorname{cosec}^2 x \) as: \[ \operatorname{cosec}^2 x = \frac{1}{\sin^2 x} \] Thus, we can rewrite the derivative as: \[ \frac{dy}{dx} = -\operatorname{cosec} x \cdot \operatorname{cosec} x \cdot \sin x \cot x \] This simplifies to: \[ \frac{dy}{dx} = -\operatorname{cosec} x \cdot \cot x \] Therefore, we have shown that: \[ \frac{dy}{dx} = -\operatorname{cosec} x \cot x \] Thus, the result is verified: \[ \frac{dy}{dx} = -\operatorname{cosec} x \cot x \] This completes the proof.