x^2/2 is 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.