Compute the derivative of the given function. \( f^{\prime}(x)=\square f(x)=x^{8} e^{x} \cot x \)

To compute the derivative of the function \( f(x) = x^8 e^x \cot x \), we will use the product rule and the chain rule. The product rule states that if you have a product of two functions \( u(x) \) and \( v(x) \), then the derivative is given by: \[ (uv)' = u'v + uv' \] In this case, we can identify three functions: - \( u(x) = x^8 \) - \( v(x) = e^x \) - \( w(x) = \cot x \) Thus, we can express \( f(x) \) as: \[ f(x) = u(x) v(x) w(x) = x^8 e^x \cot x \] To find \( f'(x) \), we will apply the product rule multiple times. First, we differentiate \( f(x) \): \[ f'(x) = (u v w)' = (u v)' w + u v w' \] Now we need to compute \( (u v)' \) and \( w' \). 1. **Differentiate \( u v \)**: Using the product rule on \( u(x) = x^8 \) and \( v(x) = e^x \): \[ (u v)' = u' v + u v' = (8x^7)(e^x) + (x^8)(e^x) = e^x (8x^7 + x^8) \] 2. **Differentiate \( w(x) = \cot x \)**: The derivative of \( \cot x \) is: \[ w' = -\csc^2 x \] Now we can substitute back into our expression for \( f'(x) \): \[ f'(x) = (u v)' w + u v w' = e^x (8x^7 + x^8) \cot x + x^8 e^x (-\csc^2 x) \] Now we can factor out \( e^x \): \[ f'(x) = e^x \left( (8x^7 + x^8) \cot x - x^8 \csc^2 x \right) \] Thus, the derivative of the function \( f(x) = x^8 e^x \cot x \) is: \[ \boxed{f'(x) = e^x \left( (8x^7 + x^8) \cot x - x^8 \csc^2 x \right)} \]