Question: Let \( f(x) = xe^x + \frac{x}{e^x} + 3\cos(x) \). Find

Solution

To solve this problem, we need to find the derivative of the function \( f(x) = xe^x + \frac{x}{e^x} + 3\cos(x) \). First, let’s differentiate each term separately. The first term is \( xe^x \). Using the product rule, which states \((uv)’ = u’v + uv’\), where \( u = x \) and \( v = e^x \), we find: \[ u’ = 1, \quad v’ = e^x \] Thus, the derivative of \( xe^x \) is: \[ (xe^x)’ = 1 \cdot e^x + x \cdot e^x = e^x + xe^x \] The second term is \( \frac{x}{e^x} \). This can be rewritten as \( x \cdot e^{-x} \). Using the product rule again with \( u = x \) and \( v = e^{-x} \), we have: \[ u’ = 1, \quad v’ = -e^{-x} \] Thus, the derivative of \( x \cdot e^{-x} \) is: \[ (x \cdot e^{-x})’ = 1 \cdot e^{-x} + x \cdot (-e^{-x}) = e^{-x} - xe^{-x} \] The third term is \( 3\cos(x) \). The derivative of \( 3\cos(x) \) is: \[ (3\cos(x))’ = -3\sin(x) \] Now, combine the derivatives of all terms: \[ f’(x) = (e^x + xe^x) + (e^{-x} - xe^{-x}) - 3\sin(x) \] Thus, the derivative of the function \( f(x) \) is: \[ f’(x) = e^x + xe^x + e^{-x} - xe^{-x} - 3\sin(x) \]