In your example, y is a function of x. x is also a function of y. If you pick a value for x, there is a unique y that satisfies the equality. If you pick a valu

In your example, y is a function of x. x is also a function of y. If you pick a value for x, there is a unique y that satisfies the equality. If you pick a value for y, there a unique x that satisfies the equality. Let's find y as a function of x x^2y = 5 => y = 5x^-2 <------ this is y(x) That's all it means. This a function, because as mentioned, x uniquely determines y. Now let's look at an example where y is not a function of x - the standard example of a circle of radius r: x^2 + y^2 = r^2 solving for y gives y = +- sqrt(r^2 - x^2). This is NOT a function, because for a value of x, we don't know which value of y to choose - it could be the positive case or the negative case. Geometrically, this means a line (like y = c) can intersect the circle in two places, so trying to find the single intersection point is not well defined. But the gradient y' IS a function, a function of both x and y and we can find it by implicit differentiation (with respect to x): 2x + 2yy' = 0 => y' = -x/y <---- this is y'(x,y) Geometrically this means that given a point (x,y) on a circle, the gradient is a well defined quantity, even though y(x) is not. ------------ Also: I had a look at your submission history and I really don't think that HN is the right place for most of your questions. I'm just procrastinating by answering this. You should probably be using math.stackexchange.com or a similar site.