Can x^2/2 + y Be Considered an Antiderivative of x?

[LucasGB]

x^2/2 is an antiderivative of x, for the derivative of x^2/2 with respect to x is x. Formally speaking, can I consider x^2/2 + y, where y is a variable and not a constant, to be an antiderivative of x, since the partial derivative of x^2/2 + y with respect to x equals x?

 

[pbandjay]

When dealing with a differentiable function f(x,y), if f_x = x then f(x,y) = x²/2 + g(y), where g is a function of y. An easy application of this kind of anti-differentiation is solving Exact Differential Equations. I don't know, that's my two cents worth.

 

[n!kofeyn]

Sure, why not? Also, why do you ask?

 

[LucasGB]

For no special reason. I'm just trying to understand if one can formally define an antiderivative of f(x) to be any function F(x,y,...,n) whose derivative with respect to x is f(x), or if an antiderivative is specifically those functions F(x) whose derivative w.r.t x is f(x).

 

[JSuarez]

Strictly speaking, no. Because the function f(x) = x and f(x,y) = x are fundamentally different. Notice that the derivative of the first is also a function of the same type, while in the second is given by a matrix.
If you consider x²/2 + y to be an antiderivative of x, then the variables x and y must be considered in equal footing, so why prefer the partial derivative relative to x? You must also admit xy as an antiderivative to x.