Partial derivatives of the function f(x,y)

[fog37]

TL;DR

Partial derivatives of the function f(x,y)

Hello,
Given a function like $z= 3x^2 +2y$, the partial derivative of z w.r.t. x is equal to: $$\frac {\partial z}{\partial x} = 6x$$

Let's consider the point $(3,2)$. If we sat on top of the point $(3,2)$ and looked straight in the positive x-direction, the slope The slope would be $(6)(2)=12$. In taking the partial derivative, we assume that y is fixed, i.e. kept constant at $y=2$. However, if we picked a different starting point like $(3,4)$ that has a different $y$ value, the partial derivative would still be equal to $$\frac {\partial z}{\partial x} =12 $$.
But I am envisioning a $z$ curve over the x-y plane that may have a different slope in the x-direction at the point $(3,4)$. That seems possible. However, $$\frac {\partial z}{\partial x} = 6x$$ does not capture the fact that the local slope in the x-direction may be different at different y locations....Where am I off?

Thanks!

 

[anuttarasammyak]

TL;DR Summary: Partial derivatives of the function f(x,y)

The slope would be (6)(2)=12.

\nabla z(x,y)=(6x,2)
\nabla z(3,2)=(18,2)
\nabla z(3,4)=(18,2)

 

[mathman]

Fix y and z is a parabola. As functions of y you have a set of parallel parabolas, so for a given x the slope is the same for all.

 

[Steve4Physics]

Can I add a couple of points to what has already been said.

Let's consider the point $(3,2)$. If we sat on top of the point $(3,2)$ and looked straight in the positive x-direction, the slope The slope would be $(6)(2)=12$.

No, The x-direction slope is $6x = 6 \times 3 = 18$, not $12$.

In taking the partial derivative, we assume that y is fixed, i.e. kept constant at $y=2$. However, if we picked a different starting point like $(3,4)$ that has a different $y$ value, the partial derivative would still be equal to $$\frac {\partial z}{\partial x} =12 $$.

You mean would still be equal to $18$, not $12$.

But I am envisioning a $z$ curve over the x-y plane

You mean a curved surface over the xy plane. For any point $(x,y)$ the 'height of the surface over that point is $z=f(x,y)$.

However, $$\frac {\partial z}{\partial x} = 6x$$ does not capture the fact that the local slope in the x-direction may be different at different y locations....Where am I off?

For this particular surface, the x-slope doesn't depend on $y$. It helps to envisage the surface. If you can't imagine it, there are various 3D plotters available, e.g. look at this: https://www.math3d.org/IGJSjfMEG

Edit.

 

[fog37]

Thank you! The surface visualization app helped a lot.

In general, given $z=f(x,y)$, the partial derivative w.r.t. $x$ can be constant, depend only on $x$, or depend on both $x$ and $y$. The partial derivative is the slope at a particular point when we make an infinitesimal step in the x-direction and no step at all in the y-direction. When we say that $\frac {\partial z} {\partial x}$ assumes that we keep $y$ fixed and at some constant level, it does not necessarily imply that that partial derivative cannot depend on $y$ itself (it does only for linear models).

Example: consider the points $(2,1)$ and $(3,1)$, and the partial derivative $$\frac {\partial z} {\partial x} = 3x+y$$ The slopes are $7$ and $10$. The $slope=7$ means that constraining/fixing $y=1$, the derivative is $\frac {\partial z} {\partial x} = 3x+1$. Keeping $y$ fixed is the same as controlling for the variable $y$: this means that we are considering the variable $y$ in our model and exploring the change in $z$ for changes in $x$ while #y# is not changing, kept constant. That is what controlling means: considering the variable and keeping it a some fixed value while the other variables change so we can isolate their contributions to the dependent variable.

In the case of multiple linear regression, we could have a model like $$y=3x_1+2x_2$$. We say that 3 is the main effect of $x_1$ when $x_2$ is kept constant because $\frac {\partial y} {partial x} = 3$. The surface is a tilted plane and the slopes in the $x_1$-direction are all the same for a a certain value of $x_1$ value regardless of the value of $x_2$.

If we don't include $x_2$ in our model, we are not controlling for $x_2$ essentially. The $\frac {\partial y} {partial x}$ remains exactly the same, $3$, even without controlling though...So what is the actual point of including $x_2$ if it really does not change the partial derivative of $z$ w.r.t. $x_1$?

Sorry for the wordiness...

 

[Steve4Physics]

Sounds like you have the right idea.

When we say that $\frac {\partial z} {\partial x}$ assumes that we keep $y$ fixed and at some constant level,

Beware of terminology and mixing-up two different things.

Keeping $y$ 'fixed' (at a constant value) is one thing.

But a 'constant level' is usually interpreted (in the present context) as a constant value of $z$, in the same way that a particular contour on a map indicates a constant height-level.

You can't do both at the same time. So it's confusing to say "we keep $y$ fixed and at some constant level".

On a more general point, for surfaces defined by $z = f(x,y)$ (or using other symbols for the variables) it often helps to visualise the surface and imagine your are on it. Then you can ask what happens to your 'height' ($z$) if you move in a particular direction.

 

[pasmith]

If your function is of the form g(x) + h(y), the partial derivative with respect to one variable will never depend on the other; if you want that then you need a more general form.