\( 1 \leftarrow \quad \) Find the first partial derivatives of the following function. \[ f(x, y)=x^{6}-6 x y-8 y^{2} \] \( f_{x}(x, y)=6 x^{5}-6 y \) \( f_{y}(x, y)=\square \)

Find the partial derivative by following steps: - step0: Differentiate: \(\frac{\partial}{\partial y}\left(x^{6}-6xy-8y^{2}\right)\) - step1: Use differentiation rules: \(\frac{\partial}{\partial y}\left(x^{6}\right)-\frac{\partial}{\partial y}\left(6xy\right)-\frac{\partial}{\partial y}\left(8y^{2}\right)\) - step2: Evaluate: \(0-6x-16y\) - step3: Remove 0: \(-6x-16y\) To find the first partial derivative of the function \( f(x, y) = x^{6} - 6xy - 8y^{2} \) with respect to \( y \), we differentiate the function while treating \( x \) as a constant. The partial derivative with respect to \( y \) is calculated as follows: \[ f_{y}(x, y) = \frac{\partial}{\partial y}(x^{6} - 6xy - 8y^{2}) = -6x - 16y \] Thus, the first partial derivative with respect to \( y \) is: \[ f_{y}(x, y) = -6x - 16y \] So, the completed derivatives are: - \( f_{x}(x, y) = 6x^{5} - 6y \) - \( f_{y}(x, y) = -6x - 16y \)