Differentiate \( \frac{y-x}{1+x y} \)

To differentiate the expression \( \frac{y-x}{1+x y} \), we can apply the quotient rule, which states that if we have a function in the form \( \frac{u}{v} \), its derivative is given by \( \frac{u'v - uv'}{v^2} \). Here, let \( u = y - x \) and \( v = 1 + xy \). First, we find the derivatives \( u' \) and \( v' \): - \( u' = \frac{dy}{dx} - 1 \) (using implicit differentiation) - \( v' = y + x \frac{dy}{dx} \) (using the product rule for \( xy \)) Now, we apply the quotient rule: \[ \frac{d}{dx} \left( \frac{y - x}{1 + xy} \right) = \frac{(u'v - uv')}{v^2} = \frac{((\frac{dy}{dx} - 1)(1 + xy) - (y - x)(y + x \frac{dy}{dx})}{(1 + xy)^2} \] This simplifies the derivative but will require careful substitution and arithmetic to fully express \( \frac{dy}{dx} \) dependent on other terms. It can be quite complex, but this approach ensures that you consider each part correctly as you differentiate!