# Partial derivatives of the function f(x,y)

• I
• fog37
In summary, the conversation discusses partial derivatives of a function and how they can depend on both x and y. It also mentions the importance of controlling for a variable in order to isolate its contribution to the dependent variable. A visualization app is recommended for better understanding of the surface.
fog37
TL;DR Summary
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!

fog37 said:
TL;DR Summary: Partial derivatives of the function f(x,y)

The slope would be (6)(2)=12.
$$\nabla z(x,y)=(6x,2)$$
$$\nabla z(3,2)=(18,2)$$
$$\nabla z(3,4)=(18,2)$$

Last edited:
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.

fog37
Can I add a couple of points to what has already been said.

fog37 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##.

fog37 said:
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##.

fog37 said:
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)##.

fog37 said:
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...

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Sounds like you have the right idea.

fog37 said:
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.

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.

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