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

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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:
- $$ f_{x}(x, y) = 6x^{5} - 6y $$
- $$ f_{y}(x, y) = -6x - 16y $$
The first partial derivative of $$ f(x, y) = x^{6} - 6xy - 8y^{2} $$ with respect to $$ y $$ is $$ f_{y}(x, y) = -6x - 16y $$.

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

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