If \( f(x)=4 \sin (x)+4 x^{x} \), find \( f^{\prime}(2) \)

To find \( f^{\prime}(2) \), we need to differentiate \( f(x) = 4 \sin(x) + 4 x^{x} \) and then evaluate the derivative at \( x = 2 \). First, let's differentiate each term individually: 1. The derivative of \( 4 \sin(x) \) is \( 4 \cos(x) \). 2. For \( 4 x^{x} \), we can use the function \( g(x) = x^x \). The derivative of \( x^x \) is found using the property of logarithmic differentiation: \[ y = x^x \implies \ln y = x \ln x \] Differentiating both sides gives: \[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \implies \frac{dy}{dx} = y (\ln x + 1) = x^x (\ln x + 1) \] So, the derivative of \( 4 x^{x} \) is: \[ 4 \frac{d}{dx}(x^x) = 4 x^x (\ln x + 1) \] Now, combining the derivatives: \[ f'(x) = 4 \cos(x) + 4 x^x (\ln x + 1) \] Next, we need to evaluate \( f'(2) \): 1. Calculate \( 4 \cos(2) \). 2. Calculate \( 4 \cdot 2^2 \cdot (\ln 2 + 1) \). Calculating each part: - \( \cos(2) \) is approximately \( -0.4161 \), so \( 4 \cos(2) \approx 4 \times -0.4161 \approx -1.6644 \). - \( 2^2 = 4 \) and \( \ln(2) \) is approximately \( 0.6931 \), so \( \ln(2) + 1 \approx 1.6931 \). - Hence, \( 4 \cdot 4 \cdot (1.6931) \approx 16 \cdot 1.6931 \approx 27.0896 \). Finally, we sum these values: \[ f'(2) \approx -1.6644 + 27.0896 \approx 25.4252 \] Thus, the value of \( f'(2) \) is approximately \( 25.4252 \).