Let \( f(x)=\frac{x^{5}(x-3)^{5}}{\left(x^{2}+7\right)^{2}} \) Use logarithmic differentiation to determine the derivative. \( f^{\prime}(x)=\square \) \( -f^{\prime}(3)=\square \)

To find the derivative of the function \( f(x)=\frac{x^{5}(x-3)^{5}}{\left(x^{2}+7\right)^{2}} \) using logarithmic differentiation, we first take the natural logarithm of both sides: \[ \ln f(x) = \ln \left( \frac{x^{5}(x-3)^{5}}{(x^{2}+7)^{2}} \right). \] Using properties of logarithms, we can rewrite this as: \[ \ln f(x) = 5 \ln x + 5 \ln (x-3) - 2 \ln (x^{2}+7). \] Now differentiate both sides with respect to \( x \): \[ \frac{f'(x)}{f(x)} = \frac{5}{x} + \frac{5}{x-3} - \frac{2(2x)}{x^{2}+7}. \] Multiplying through by \( f(x) \) gives: \[ f'(x) = f(x) \left( \frac{5}{x} + \frac{5}{x-3} - \frac{4x}{x^{2}+7} \right). \] Next, substituting back \( f(x) \): \[ f'(x) = \frac{x^{5}(x-3)^{5}}{(x^{2}+7)^{2}} \left( \frac{5}{x} + \frac{5}{x-3} - \frac{4x}{x^{2}+7} \right). \] To find \( -f'(3) \), we will plug \( x = 3 \) into the derivative. Notice that \( f(3) = \frac{3^{5}(3-3)^{5}}{(3^{2}+7)^{2}} = 0 \) because of the \( (x-3)^{5} \) term, therefore \( f'(3) = 0 \) as well. Consequently, \[ -f'(3) = 0. \] So the answers are: \[ f'(x)= \frac{x^{5}(x-3)^{5}}{(x^{2}+7)^{2}} \left( \frac{5}{x} + \frac{5}{x-3} - \frac{4x}{x^{2}+7} \right) \] \[ -f'(3)= 0. \]