Partial derivatives

Calculus Level 3

True or False

Let $f \colon \mathbb {R}^2 \to \mathbb {R}$ be a continuous function.

Then the derivative $\dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial y}\right)$ can exist without $\dfrac{\partial f}{\partial x}$ existing.

True False

2 solutions

Hjalmar Orellana Soto
Jun 21, 2017

For any pair of functions $f$ and $g$ such that $F(x,y)=f(x)+g(y)$ and $g$ is $C^1$ class, we have: $f'(x)\nexists \Rightarrow \frac{\partial F}{\partial x} \nexists$ $\frac{\partial F}{\partial y}=g'(y) \Rightarrow \frac{\partial^2 F}{\partial x \partial y} =\frac{\partial }{\partial x}g'(y)=0 \exists$

Did you mean F(x,y) (uppercase) in some of the derivatives?

Leonardo Lessa - 3 years, 11 months ago

ooh, yes, let's edit it xD

Hjalmar Orellana Soto - 3 years, 11 months ago

I think you should mention that $f$ is of class less than $C^1$

Agnishom Chattopadhyay - 3 years, 11 months ago

not necessary for this, as tou can see I supposed that $f'(x)\neq$

Hjalmar Orellana Soto - 3 years, 11 months ago

sorry, I supposed that $f'(x) \nexists$

Hjalmar Orellana Soto - 3 years, 11 months ago

@Hjalmar Orellana Soto – Okay, sure. That is also another valid way to say it.

Agnishom Chattopadhyay - 3 years, 11 months ago

Abhishek Sinha
Jul 1, 2017

A simple example of such a function is $f(x,y)=|x|.$ We have $\frac{\partial f}{\partial y}=0$ , and hence, $\frac{\partial}{\partial x}(\frac{\partial f}{\partial y})=0$ everywhere, whereas $\frac{\partial f}{\partial x}$ does not exists at $x=0$ .

Partial derivatives

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