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### Course: Multivariable calculus>Unit 2

Lesson 1: Partial derivatives

# Symmetry of second partial derivatives

There are many ways to take a "second partial derivative", but some of them secretly turn out to be the same thing.  Created by Grant Sanderson.

## Want to join the conversation?

• is there no second partial derivative with respecte to y? So: fyy(x.y)
• Just a minor correction: What the previous reply meant to say was either "He just didn't write it down," or "He just forgot to write it down."

At least I hope that is what the person meant to say :)
• Is taking a partial derivative a commutative operation over the set of all real valued continuous differentiable multivariable functions?
• I would like to point out, that operands of taking a partial derivative are
-a function
-a variable that you are taking a derivative with respect to
So, using somewhat ugly infix notation (when ' * ' stands for derivative) it would look like:
f * x is the partial derivative of f with respect to x
We have (f * x) * y = (f * y) * x
But is f * x = x * f ? Certainly not.
• What's the intuition/meaning for taking partial derivative with respect to x first, then taking second partial derivative with respect to y after that? I mean the first step we already consider y term constant, so the second step seems to be inconsistent with the first one?
• I'm late, but d^2f/dydx is essentially saying how the rate of change of x changes as you move in the y direction.
So if at a point the rate of change of x is 2, but when you move in the y direction a little bit the rate of change of x is 1, then d^2f/dydx is negative.
• Isn't the denominator supposed to be partial squared times x^2?
• when you take a second derivative and are using Leibniz notation, think of it as the 'd's in the numerator getting squared and the 'dx's in the denominator being squared.

So d/dx(dy/dx)= d*dy / dx*dx = d^2y/dx^2
• I don't get exactly what happens at . It seems kind of chain rule, but since it's considering cos(x) as constant, wouldn't its derivative be 0 ?
(1 vote)
• In this case, cos(x) is a constant which is coefficient to your variable y^2. If you were to derive 3x with respect to x, you would get 3. Similarly, deriving (y^2)cos(x) with respect to y would keep your constant (cos(x)) as is, while taking the derivative of your variable (y^2 becomes 2y).
• Maybe I am just bad at math, but I thought the derivative of a constant was supposed to be zero... Why in the beginning of the video did he leave the constants after taking the derivative instead of zeroing them out?
• I don't think he does that. If you are asking why he doesn't touch the part of the expression that belongs to the other variable, then he does that because that is how you take a partial derivative. You only differentiate the part of the expression that contains the variable you are taking the partial derivative of.
• What if you went into third, fourth, fifth, etc partial derivatives after the end of the video?
• Yes you could. You'd just keep performing the differentiation with respect to your chosen variable, as many time as you want. So you could calculate something like fxxyxxyyyx if you had a need to. If you're interested, take f(x,y)=sin(x)e^(2y) and derive fxxyy, then fxyyx and see what happens.
• at , why the second derivative written like this (in the Numerator : the superscript of 2 is written after the partial derivative symbol, while in the denominator it is written after the x ).. what is the intuition of this writing ?
• when you take a second derivative and are using Leibniz notation, think of it as the 'd's in the numerator getting squared and the 'dx's in the denominator being squared.

So d/dx(dy/dx)= d*dy / dx*dx = d^2y/dx^2