Question on Partial Derviatives For function $f(x,y) = x^2 y$
The partial derivatives for $x$ is $2.x.y $.
I'm new to such math equation and i'm learning them now. 
May i know why is it so?
Thanks!
 A: It is because you are treating $y$ as a constant. The derivative of $y$ with respect to $x$ is zero if $y$ is not a function of $x$. Therefore, by the chain rule
$$\frac{\partial}{\partial x} (x^2 y) = x^2 \frac{\partial y}{\partial x} + y\frac{\partial}{\partial x} x^2 = x^2 \cdot 0 + y \cdot 2x = 2xy.$$
You can think of $y$ as a constant because, by definition, the operator $\partial/\partial x$ acts only on functions of $x$, not $y$. Thus,
$$\frac{\partial}{\partial x} (x^2 y) = y \frac{d}{dx} x^2 = 2xy.$$
A: The whole point of partial derivatives is that we hold all variables constant except the one we want to know the rate of change of. You can think of this geometrically in three dimensions as follows: The partial derivative defines a curve that's traced along the tangent plane to a surface z = f (x,y) where at the point in the domain where y= a, the partial derivative defines a tangent line in the domain to the point a. Here's a simple example: let y =1 for every x in the domain. Then the partial derivative defines the straight line in the domain z= 2x for every ordered pair (x,1) in the domain the plane. In reality, it's not really that simple-you have to worry about delicate continuity and limit conditions and to really make the geometry precise even in 3 dimensioms, you need projection functions. But you should be able to draw a picture now that should clarify things!    
