Для задачи на условный экстремум f(x, y) = x<sup>2</sup> – y<sup>2</sup>, φ(x, y) = x<sup>2</sup> + y<sup>2</sup> – 2 = 0 дифференциал функции связи равен: 2dx<sub>1</sub><sup>2</sup> + 2dx<sub>2</sub><sup>2</sup>, 2x<sub>1</sub><em>d</em>x<sub>1</sub> + 2x<sub>2</sub><em>d</em>x<sub>2</sub>, dx<sub>1</sub><sup>2</sup> + dx<sub>2</sub><sup>2</sup>, x<sub>1</sub><em>d</em>x<sub>1</sub> + x<sub>2</sub><em>d</em>x<sub>2</sub>