To differentiate the given function [tex]\( f(x, y) = \frac{1}{x^2 - 12} \)[/tex] with respect to [tex]\( x \)[/tex] and [tex]\( y \)[/tex], let's go through the process step-by-step:
### Differentiation with respect to [tex]\( x \)[/tex]:
First, recognize that [tex]\( y \)[/tex] does not appear in [tex]\( f(x, y) \)[/tex], so differentiation with respect to [tex]\( x \)[/tex] will be treated as if [tex]\( f \)[/tex] is a function of [tex]\( x \)[/tex] alone.
Given:
[tex]\[ f(x, y) = \frac{1}{x^2 - 12} \][/tex]
We need to find [tex]\(\frac{\partial f}{\partial x}\)[/tex]. The function [tex]\( f(x, y) \)[/tex] can be rewritten as:
[tex]\[ f(x, y) = (x^2 - 12)^{-1} \][/tex]
Using the chain rule and the power rule for differentiation, we get:
[tex]\[ \frac{\partial f}{\partial x} = \frac{d}{dx}\left((x^2 - 12)^{-1}\right) \][/tex]
Let [tex]\( u = x^2 - 12 \)[/tex], hence [tex]\( f = u^{-1} \)[/tex].
By the chain rule:
[tex]\[ \frac{d}{dx}\left(u^{-1}\right) = -u^{-2} \cdot \frac{du}{dx} \][/tex]
Differentiate [tex]\( u = x^2 - 12 \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{du}{dx} = 2x \][/tex]
So,
[tex]\[ \frac{d}{dx}\left((x^2 - 12)^{-1}\right) = - (x^2 - 12)^{-2} \cdot 2x \][/tex]
Thus,
[tex]\[ \frac{\partial f}{\partial x} = - \frac{2x}{(x^2 - 12)^2} \][/tex]
### Differentiation with respect to [tex]\( y \)[/tex]:
Since [tex]\( f(x, y) = \frac{1}{x^2 - 12} \)[/tex] does not contain [tex]\( y \)[/tex], the partial derivative with respect to [tex]\( y \)[/tex] is straightforward.
[tex]\[ \frac{\partial f}{\partial y} = 0 \][/tex]
### Summary:
The partial derivatives of the given function are:
[tex]\[ \frac{\partial f}{\partial x} = -\frac{2x}{(x^2 - 12)^2} \][/tex]
[tex]\[ \frac{\partial f}{\partial y} = 0 \][/tex]
Therefore, the solution in a compact form is:
[tex]\[ \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = \left( -\frac{2x}{(x^2 - 12)^2}, 0 \right). \][/tex]
