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

To find the derivative of the function \( \frac{y-x}{1+xy} \) with respect to \( x \), we will use the quotient rule. The quotient rule states that if you have a function \( \frac{u}{v} \), then its derivative is given by \( \frac{u'v - uv'}{v^2} \). Let \( u = y - x \) and \( v = 1 + xy \). First, we calculate the derivatives: - \( u' = \frac{dy}{dx} - 1 \) (since the derivative of \( -x \) is \( -1 \)) - \( v' = y + x \frac{dy}{dx} \) (using the product rule for \( xy \)) Now we apply the quotient rule: \[ \frac{d}{dx} \left( \frac{y-x}{1+xy} \right) = \frac{(u'v - uv')}{v^2} \] Substituting in our derivatives: \[ = \frac{((\frac{dy}{dx} - 1)(1 + xy) - (y - x)(y + x \frac{dy}{dx}))}{(1 + xy)^2} \] This derivative expression can be simplified further based on specific contexts or by plugging in known values, but this gives you the formula to work from.