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What is the derivative with respect to x [math]\frac{x}{y} + \frac{y}{x} =1 [/math]?
Can you walk me through the process of finding the derivative?
Sure! So, if we differentiate both sides of the equation [math]\frac{x}{y} + \frac{y}{x} =1[/math] with respect to x, we get:

[math]\frac{d}{dx}\left(\frac{x}{y} + \frac{y}{x}\right) = \frac{d}{dx}(1)[/math]

Using the product rule, we can find the derivative of the left-hand side:

[math]\frac{d}{dx}\left(\frac{x}{y} + \frac{y}{x}\right) = \frac{d}{dx}\left(\frac{x}{y}\right) + \frac{d}{dx}\left(\frac{y}{x}\right)[/math]
[math]= \frac{y\frac{d}{dx}(x) - x\frac{d}{dx}(y)}{y^2} + \frac{x\frac{d}{dx}(y) - y\frac{d}{dx}(x)}{x^2}[/math]
[math]= \frac{1}{y} - \frac{x}{y^2}\frac{dy}{dx} + \frac{1}{x}- \frac{y}{x^2}\frac{dy}{dx}[/math]

Simplifying the right-hand side, we get:

[math]\frac{d}{dx}(1) = 0[/math]

Putting it altogether, we have:

[math]\frac{1}{y} - \frac{x}{y^2}\frac{dy}{dx} + \frac{1}{x}- \frac{y}{x^2}\frac{dy}{dx} = 0[/math]

Now we just need to solve for [math]\frac{dy}{dx}[/math]:

[math]\frac{x}{y^2}\frac{dy}{dx} + \frac{y}{x^2}\frac{dy}{dx} = \frac{1}{y} + \frac{1}{x}[/math]
[math]\frac{dy}{dx}\left(\frac{x}{y^2} + \frac{y}{x^2}\right) = \frac{x+y}{xy}[/math]
[math]\frac{dy}{dx}=\frac{(x+y)xy}{(xy)^2}\left(\frac{x}{y^2} + \frac{y}{x^2}\right)^{-1}[/math]