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 \( \frac{d y}{d x} = -\operatorname{cosec} x \cot x \), we need to differentiate \( y \) with respect to \( x \). First, recall the definition of the cosecant function: \[ \operatorname{cosec} x = \frac{1}{\sin x} \] Now, we will differentiate \( y \) with respect to \( x \) using the chain rule. The chain rule states that if you have a composite function \( f(g(x)) \), then the derivative of \( f \) with respect to \( x \) is \( f'(g(x))g'(x) \). Let's differentiate \( y = \operatorname{cosec} x \) with respect to \( x \): \[ \frac{d y}{d x} = \frac{d}{d x} \left( \frac{1}{\sin x} \right) \] To differentiate \( \frac{1}{\sin x} \), we can use the quotient rule, which states that if you have a function \( f(x) = \frac{g(x)}{h(x)} \), then the derivative of \( f \) with respect to \( x \) is given by: \[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \] In our case, \( g(x) = 1 \) and \( h(x) = \sin x \). The derivative of \( g(x) \) is \( g'(x) = 0 \) because the derivative of a constant is zero. The derivative of \( h(x) \) is \( h'(x) = \cos x \). Applying the quotient rule: \[ \frac{d y}{d x} = \frac{0 \cdot \sin x - 1 \cdot \cos x}{[\sin x]^2} \] \[ \frac{d y}{d x} = \frac{-\cos x}{\sin^2 x} \] Now, we can rewrite this in terms of the cotangent function, which is defined as \( \cot x = \frac{\cos x}{\sin x} \): \[ \frac{d y}{d x} = -\frac{\cos x}{\sin^2 x} = -\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = -\operatorname{cosec} x \cot x \] Thus, we have shown that if \( y = \operatorname{cosec} x \), then \( \frac{d y}{d x} = -\operatorname{cosec} x \cot x \).