Originally Posted by

**paolopiace** I encountered the following implication in a text book:

$\displaystyle

g(x,y) = - \frac{\partial}{\partial y} f(x,y)\ \Longrightarrow\

\underbrace{\frac{\partial}{\partial y} \frac{1}{2} f^2(x,y)}_{(a)} =

g(x,y) \int_x^y g(x,u) du

$

which I'm strugling to understand...

Being (a) = f(x,y), it can be derived integrating the first equation:

$\displaystyle

\int_x^y g(x,u) du = \int_x^y -\frac{\partial}{\partial u} f(x,u) = -f(x,u) \vert_x^y = -f(x,y)

$

That's because f(x,x) := 0 by definition.

Therefore the initial implication does not make sense to me.

Does it sense to some you guys?

If yes, would you pls help me to understand/derive it?

Thanks!