1. **Problem:** Compute the Jacobian \( \frac{\partial(u,v)}{\partial(x,y)} \) where \( u = x + y \) and \( v = xy + 1 \). 2. **Formula:** The Jacobian determinant for functions \( u(x,y) \) and \( v(x,y) \) is given by: $$ J = \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} = \frac{\partial u}{\partial x} \cdot \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y} \cdot \frac{\partial v}{\partial x} $$ 3. **Calculate partial derivatives:** - \( \frac{\partial u}{\partial x} = 1 \) because \( u = x + y \) - \( \frac{\partial u}{\partial y} = 1 \) - \( \frac{\partial v}{\partial x} = y \) because \( v = xy + 1 \) - \( \frac{\partial v}{\partial y} = x \) 4. **Substitute into Jacobian:** $$ J = (1)(x) - (1)(y) = x - y $$ 5. **Interpretation:** The Jacobian \( J = x - y \) depends on \( x \) and \( y \). It is zero when \( x = y \). **Final answer:** $$ \boxed{\frac{\partial(u,v)}{\partial(x,y)} = x - y} $$