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## Multivariable calculus

### Course: Multivariable calculus>Unit 2

Lesson 1: Partial derivatives

# Graphical understanding of partial derivatives

One of the best ways to think about partial derivatives is by slicing the graph of a multivariable function.  Created by Grant Sanderson.

## Want to join the conversation?

• If the derivative of a constant*variable = constant how come in the first evaluation the partial derivative respect to x =>x²*y=2xy and in the second evaluation the partial derivative respect to y=>x²*y=x². I know that the power rule but don't understand why the place of the constant matters.
• But the place of the constant doesn't matter. In the first evaluation of partial derivative respect to x => x^2y = 2xy because we are considering y as constant, therefore you may replace y as some trivial number a, and x as variable, therefore derivative of x^2y is equivalent to derivative of x^2.a which is 2a.x , substitute trivial a with y and we have 2xy. In the latter case we considering x as trivial a and y as constant so we get 1.x^2
• I understand the video， but for only 2 variables，how can you have 3-dimention image
• not sure exactly what you mean but i'll try and answer.

If you mean you don't understand why we have a 3d image for a problem with 2d input it's because the 3rd dimension represents your output.

If you mean you get how this works for 2 variables but like to see how it looks for 3, we can't really graph that since our brain can't perceive a fourth dimension. I think this is what he was getting at, the graphical intuition is good but gets you only thus far with multivariable calculus because of the high dimensionality. There is still ways to represent such functions graphically, plotting input and output separately for example is what comes to mind, but I am just starting to learn calculus, so someone more advanced needs to answer.
• The equation in the video using only x and y but the graph also has the z-axis. So can we assume that the z was missing as part of the equation?
• what software are you using to make the plot? what about the plane?
• I'm having hard time to visualize the partial derivative of a sphere,
if we for example took the y-axis as our constant plane, then we would have two different (z) values for a given (x) value, each of which has different slope (derivative).
As a representation for this part it would look like a circle, so the two different (z) values would be as follows : the upper value and the lower value; and they will have slopes that are opposite to each other in signs.
Hence, our partial derivative function should return 2 different answers, one for the upper (z) value and one for the lower (z) value. But this simply doesn't happen, can someone explain to me. It would be great if there were a graphical representation of this situation.

As you said ,the part would look like a circle
Let's assume the equation of the sphere as z^2 +x^2+y^2=25
and let our constant y plane be y=2

So the circle's equation will be
z^2=25-x^2-(2)^2
z^2=21-x^2

Now the equation of the upper half of the circle will be
z=+sqrt(21-x^2)
and lower part of the circle will be
z=-sqrt(21-x^2)

So accordingly we use these two equations

So it doesn't return 2 different answers
Hope it helped a bit
• I'm confused about the pronunciations of multi, because if it's multivariable, people pronunce multi as mult-ee. But if it's multidemensional@, he pronunced it as mult-eye.
(1 vote)
• Pronunciations vary from person to person and region to region. For multi, both versions are acceptable. I've heard it both ways as well.
• how do i know that this is the 3d graph i get before starting to solve the problem?
• the partial derivative with respect to X equals 2XY
how come the graph is a shifted square At ?? shouldn't it be a straight line because Y is const?
and although the derivative equals a negative number (-2)
the point was in the positive Z axis
and at the same point the Y partial derivative gives a totally different value 1.9
then how did we draw this 3D graph?
(1 vote)
• You are correct, that the partial derivative with respect to x is 2xy, and that y is a constant. The plane (the cross section) of the graph at is JUST how the function f, looks like, when y = +1, right? And we are saying, that indeed, the derivative of that parabolic looking cross section when y = 1 is 2xy. At , that 2xy shows up. Right? So that parabola is just THE FUNCTION F, when y = 1, and NOT the derivative. When we take the derivative OF THAT CROSS SECTION, we get 2xy, just to emphasize. Have a nice day ); ^^