# Thread: partial derivative

1. ## partial derivative

I encountered the following implication in a text book:

$
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:

$
\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!

2. Hello,

Originally Posted by paolopiace
I encountered the following implication in a text book:

$
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:

$
\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!
Well, $\frac{\partial}{\partial y} \frac{1}{2} f^2(x,y){\color{magenta}=}\frac 12 \cdot \ \left(2 \cdot {\color{red}\frac{\partial}{\partial y}} {\color{blue}f(x,y)}\right) \cdot f(x,y)$ by using the chain rule for differentiation.

${\color{magenta}=}{\color{red}-g(x,y)} \cdot {\color{blue}f(x,y)}$

You have proved that $\int_x^y g(x,u) du=-f(x,y)$ --> ${\color{blue}f(x,y)}=-\int_x^y g(x,u) du$

${\color{magenta}=}-g(x,y) \cdot (-\int_x^y g(x,u) du)=g(x,y) \cdot \int_x^y g(x,u) du$