Differentiate the function \( L(x)=-2 e^{x}(\csc x+7 \cot x) \)

To find the derivative of the function \( L(x)=-2 e^{x}(\csc x+7 \cot x) \), we will need to apply the product rule and the chain rule. 1. First, identify the two parts of the product: - \( u = -2 e^{x} \) - \( v = \csc x + 7 \cot x \) 2. Differentiate \( u \) and \( v \): - The derivative of \( u \): \( u' = -2 e^{x} \) - The derivative of \( v \): To differentiate \( v \): - \( \frac{d}{dx}(\csc x) = -\csc x \cot x \) - \( \frac{d}{dx}(7 \cot x) = -7 \csc^2 x \) Therefore, \( v' = -\csc x \cot x - 7 \csc^2 x \) 3. Apply the product rule: \[ L'(x) = u'v + uv' \] Substituting \( u, u', v, v' \): \[ L'(x) = (-2 e^{x})(\csc x + 7 \cot x) + (-2 e^{x})(-\csc x \cot x - 7 \csc^2 x) \] 4. Simplifying gives: \[ L'(x) = -2 e^{x}(\csc x + 7 \cot x) + 2 e^{x}(\csc x \cot x + 7 \csc^2 x) \] 5. Combine like terms: \[ L'(x) = -2 e^{x} \csc x + 2 e^{x} \csc x \cot x - 14 e^{x} \cot x + 14 e^{x} \csc^2 x \] So the final answer is: \[ L'(x) = 2 e^{x}(\csc x \cot x - 7 \cot x + 7 \csc^2 x - \csc x) \]