If \( y=\frac{\operatorname{cosec}\left(e^{x}\right)}{e^{x}} \) then \( \frac{d y}{d x}= \)

To find \( \frac{d y}{d x} \) for \( y=\frac{\operatorname{cosec}\left(e^{x}\right)}{e^{x}} \), we will use the quotient rule. The quotient rule states that if \( y = \frac{u}{v} \), then \( \frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^2} \) where \( u = \operatorname{cosec}(e^x) \) and \( v = e^x \). First, we compute \( u \) and \( v \): - \( u = \operatorname{cosec}(e^x) = \frac{1}{\sin(e^x)} \) - \( v = e^x \) Next, we find the derivatives: - \( \frac{d v}{d x} = e^x \) - To differentiate \( u \), we use the chain rule: \[ \frac{d u}{d x} = -\operatorname{cosec}(e^x)\cot(e^x) \cdot \frac{d}{d x}(e^x) = -\operatorname{cosec}(e^x)\cot(e^x) \cdot e^x \] Now applying the quotient rule: \[ \frac{d y}{d x} = \frac{e^x \left(-\operatorname{cosec}(e^x)\cot(e^x) \cdot e^x\right) - \operatorname{cosec}(e^x) \cdot e^x}{(e^x)^2} \] This simplifies to: \[ \frac{d y}{d x} = \frac{-e^{2x}\operatorname{cosec}(e^x)\cot(e^x) - e^x \operatorname{cosec}(e^x)}{e^{2x}} \] Further simplifying gives: \[ \frac{d y}{d x} = -\operatorname{cosec}(e^x)\left(\cot(e^x) + 1\right) \] Thus, the final answer is: \[ \frac{d y}{d x} = -\operatorname{cosec}(e^x)(\cot(e^x) + 1) \]