Determine \( f_{x} \) und fy of the given function \( f(x, y)=\left(x^{5} \ln 2 x\right) y^{8} \)

To find the partial derivatives \( f_x \) and \( f_y \) of the function \[ f(x, y) = \left(x^{5} \ln(2x)\right) y^{8}, \] we will differentiate with respect to \( x \) and \( y \) separately while treating the other variable as a constant. ### Partial Derivative with Respect to \( x \) (\( f_x \)) \[ f(x, y) = x^{5} \ln(2x) \cdot y^{8} \] 1. **Differentiate \( x^{5} \ln(2x) \) with respect to \( x \):** Use the product rule: \[ \frac{d}{dx}\left(x^{5} \ln(2x)\right) = \frac{d}{dx}(x^{5}) \cdot \ln(2x) + x^{5} \cdot \frac{d}{dx}(\ln(2x)) \] Calculate each part: \[ \frac{d}{dx}(x^{5}) = 5x^{4} \] \[ \frac{d}{dx}(\ln(2x)) = \frac{1}{2x} \cdot 2 = \frac{1}{x} \] Combine them: \[ 5x^{4} \ln(2x) + x^{5} \cdot \frac{1}{x} = 5x^{4} \ln(2x) + x^{4} = x^{4}(5 \ln(2x) + 1) \] 2. **Multiply by \( y^{8} \) (since \( y \) is treated as a constant):** \[ f_x = x^{4}(5 \ln(2x) + 1) y^{8} \] ### Partial Derivative with Respect to \( y \) (\( f_y \)) \[ f(x, y) = x^{5} \ln(2x) \cdot y^{8} \] 1. **Differentiate \( y^{8} \) with respect to \( y \):** \[ \frac{d}{dy}(y^{8}) = 8y^{7} \] 2. **Multiply by \( x^{5} \ln(2x) \) (since \( x \) is treated as a constant):** \[ f_y = 8x^{5} \ln(2x) \cdot y^{7} \] ### Summary The partial derivatives of the function \( f(x, y) = \left(x^{5} \ln(2x)\right) y^{8} \) are: \[ f_x = x^{4} \left(5 \ln(2x) + 1\right) y^{8} \] \[ f_y = 8x^{5} \ln(2x) \cdot y^{7} \] **Final Answer:** After simplifying, the partial derivatives are \[ f_x = x^{4} \big(5\, \ln(2x) + 1\big)\, y^{8} \] and \[ f_y = 8\, x^{5} \ln(2x)\, y^{7}. \]