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Question

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

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To find the first partial derivatives of the function $$ f(x, y) = x^{6} - 6xy - 8y^{2} $$, we need to find the partial derivative of $$ f $$ with respect to $$ x $$ and the partial derivative of $$ f $$ with respect to $$ y $$.

The partial derivative of $$ f $$ with respect to $$ x $$ is denoted as $$ f_{x}(x, y) $$ and is found by differentiating $$ f $$ with respect to $$ x $$ while treating $$ y $$ as a constant.

Similarly, the partial derivative of $$ f $$ with respect to $$ y $$ is denoted as $$ f_{y}(x, y) $$ and is found by differentiating $$ f $$ with respect to $$ y $$ while treating $$ x $$ as a constant.

Let's find the partial derivative of $$ f $$ with respect to $$ x $$ first.
Find the partial derivative by following steps:
- step0: Differentiate:
  $$\frac{\partial}{\partial x}\left(x^{6}-6xy-8y^{2}\right)$$
- step1: Use differentiation rules:
  $$\frac{\partial}{\partial x}\left(x^{6}\right)-\frac{\partial}{\partial x}\left(6xy\right)-\frac{\partial}{\partial x}\left(8y^{2}\right)$$
- step2: Evaluate:
  $$6x^{5}-6y-0$$
- step3: Remove 0:
  $$6x^{5}-6y$$
The partial derivative of $$ f(x, y) = x^{6} - 6xy - 8y^{2} $$ with respect to $$ x $$ is $$ f_{x}(x, y) = 6x^{5} - 6y $$.
$$ f_{x}(x, y) = 6x^{5} - 6y $$

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

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