Find the linear approximation at (x, 0). f( x, y) = vy + cos2(x) = 1 + zy Let f(x, y) = vy + cos2(x). Then fx(x, y) = y+ cos-(x) and fy(x, y) = Both fx and fy are continuous functions for y so fis differentiable at (, 0) by this theorem. We have fx(x, 0) = and X fy(, 0) = 1/2 , so the linear approximation of fat (x, 0) is f(x, y) = f(, 0) + fx(n, 0)(x - *) + fy(x, 0)(y - 0) = 1 X Need Help? Read It