Is Differentiability at the Origin Determined by Partial Derivatives?

In summary, the conversation discusses the differentiability of a function $f$ defined by $f(x,y)=yg(x)$ where $g$ is an arbitrary function. It is stated that $f$ is differentiable at the origin if and only if $g$ is continuous at $x=0$. The conversation then delves into verifying the differentiability of $f$ using a specific set of conditions. It is also mentioned that the third condition of the definition of differentiability must hold, and that $g(0)$ must exist for $f$ to be differentiable. The conversation then concludes with a discussion on proving the converse using polar coordinates.
  • #1
mathmari
Gold Member
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Hey! :eek:

Let $g:\mathbb{R}\rightarrow \mathbb{R}$ be arbitrary and $f:\mathbb{R}^2\rightarrow \mathbb{R}$ be defined by $f(x,y)=yg(x)$.
I want to prove that $f$ is differentiable in the origin if and only if $g$ is continuous in $x=0$.

So that $f$ is differentiable in $(0,0)$ does the following has to hold?
  • $f_x(0,0)$ exists
  • $f_y(0,0)$ exists
  • $\displaystyle{\lim_{(x,y)\rightarrow (0,0)}\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}=0}$

Or do we have to check in an other way the differentiability? (Wondering)
 
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  • #2
Hey mathmari! (Wave)

Yes, that is the definition that we have to verify. (Nod)
 
  • #3
I like Serena said:
Yes, that is the defintion that we have to verify. (Nod)

So, we have the following:
  • The partial derivative is $f_x(x,y)=yg'(x)$. But $g$ is an arbitrary function. Can we just take its derivative? Or do we have to take this as a condition?

    Then $f_x(0,0)$ is equal to $0\cdot g'(0)$. Here it has to hold that $g'(0)$ exists, or not? (Wondering)
  • The partial derivative is $f_y(x,y)=g(x)$. At the point $(0,0)$ the partial derivative is equal to $f_y(0,0)=g(0)$. So $g(0)$ has to exist.
  • We have the following limit
    \begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-0\cdot g(0)-0\cdot g'(0)\cdot x-g(0)\cdot y}{\|(x,y)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot \left [g(x)-g(0)\right ]}{\sqrt{x^2+y^2}}\end{align*} right? How could we continue? (Wondering)
 
  • #4
mathmari said:
So, we have the following:
  • The partial derivative is $f_x(x,y)=yg'(x)$. But $g$ is an arbitrary function. Can we just take its derivative? Or do we have to take this as a condition?

    Then $f_x(0,0)$ is equal to $0\cdot g'(0)$. Here it has to hold that $g'(0)$ exists, or not? (Wondering)

We can't make assumptions about $g$ can we?
As yet we don't even know if $g$ is continuous in $0$, which is what we want to verify.
Instead we should deduce the value of $f_x(0,0)$ from the limit.

Suppose we take the path given by $(x(t),y(t))=(t,0)$.
Then the limit becomes:
$$\lim_{t\to 0} \frac{0\cdot g(t) - f_x(0,0)\cdot t - f_y(0,0)\cdot 0}{\sqrt{t^2+0^2}} = \lim_{t\to 0} \frac{-f_x(0,0)\cdot t}{|t|} = 0$$
Therefore we must have that $f_x(0,0)=0$ don't we? (Wondering)

mathmari said:
  • The partial derivative is $f_y(x,y)=g(x)$. At the point $(0,0)$ the partial derivative is equal to $f_y(0,0)=g(0)$. So $g(0)$ has to exist.

It's already given that $g(0)$ exists, isn't it?
And yes, we must have $f_y(0,0)=g(0)$.

mathmari said:
  • We have the following limit
    \begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-0\cdot g(0)-0\cdot g'(0)\cdot x-g(0)\cdot y}{\|(x,y)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot \left [g(x)-g(0)\right ]}{\sqrt{x^2+y^2}}\end{align*} right? How could we continue? (Wondering)
We can't use $g'(0)$.
However, if instead we substitute the $f_x(0,0)=0$ that we found before, we'll find the same result.

Suppose we pick the path given by $(x(t),y(t))=(t,t)$, what will we get? (Wondering)
 
  • #5
I like Serena said:
We can't make assumptions about $g$ can we?
As yet we don't even know if $g$ is continuous in $0$, which is what we want to verify.
Instead we should deduce the value of $f_x(0,0)$ from the limit.

Suppose we take the path given by $(x(t),y(t))=(t,0)$.

Do we consider that path because then the term $f_y(0,0)$ would vanish at the limit? (Wondering)
I like Serena said:
Then the limit becomes:
$$\lim_{t\to 0} \frac{0\cdot g(t) - f_x(0,0)\cdot t - f_y(0,0)\cdot 0}{\sqrt{t^2+0^2}} = \lim_{t\to 0} \frac{-f_x(0,0)\cdot t}{|t|} = 0$$
Therefore we must have that $f_x(0,0)=0$ don't we? (Wondering)

The third condition of the definition of differentiabilty must hold, i.e. the limit must be equal to $0$ and so since $\displaystyle{\lim_{t\to 0} \frac{-f_x(0,0)\cdot t}{|t|}=\lim_{t\to 0} -f_x(0,0)\cdot \text{sgn}(t)}$ it follows that $f_x(0,0)$ must be equal to $0$.
Is this correct? (Wondering)
I like Serena said:
It's already given that $g(0)$ exists, isn't it?

How do we know that this exists? I got stuck right now. (Wondering)
I like Serena said:
Suppose we pick the path given by $(x(t),y(t))=(t,t)$, what will we get? (Wondering)

We have the following:
\begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-0\cdot g(0)-0\cdot x-g(0)\cdot y}{\|(x,y)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-g(0)\cdot y}{\sqrt{x^2+y^2}}\\ & =\lim_{t\rightarrow 0}\frac{t\cdot g(t)-g(0)\cdot t}{\sqrt{t^2+t^2}} \\ & = \lim_{t\rightarrow 0}\frac{t\cdot g(t)-g(0)\cdot t}{\sqrt{2t^2}} \\ & = \lim_{t\rightarrow 0}\frac{t\cdot [g(t)-g(0)]}{\sqrt{2}\cdot |t|}\\ & = \frac{\text{sgn}(t)}{\sqrt{2}}\cdot \lim_{t\rightarrow 0}(g(t)-g(0))\end{align*} Since the limit must be equal to $0$ it must holt that $\displaystyle{\lim_{t\rightarrow 0}g(t)=g(0)}$, which means that $g$ is continuous at $0$.

Is everything correct? (Wondering)
 
  • #6
mathmari said:
Do we consider that path because then the term $f_y(0,0)$ would vanish at the limit?

Yep. (Nod)

mathmari said:
The third condition of the definition of differentiabilty must hold, i.e. the limit must be equal to $0$ and so since $\displaystyle{\lim_{t\to 0} \frac{-f_x(0,0)\cdot t}{|t|}=\lim_{t\to 0} -f_x(0,0)\cdot \text{sgn}(t)}$ it follows that $f_x(0,0)$ must be equal to $0$.
Is this correct?

Indeed.

mathmari said:
How do we know that this exists? I got stuck right now.

It's given that $g$ is a function $\mathbb R \to \mathbb R$.
That means that $g$ is defined for every real number isn't it? (Wondering)

mathmari said:
We have the following:
\begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-0\cdot g(0)-0\cdot x-g(0)\cdot y}{\|(x,y)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-g(0)\cdot y}{\sqrt{x^2+y^2}}\\ & =\lim_{t\rightarrow 0}\frac{t\cdot g(t)-g(0)\cdot t}{\sqrt{t^2+t^2}} \\ & = \lim_{t\rightarrow 0}\frac{t\cdot g(t)-g(0)\cdot t}{\sqrt{2t^2}} \\ & = \lim_{t\rightarrow 0}\frac{t\cdot [g(t)-g(0)]}{\sqrt{2}\cdot |t|}\\ & = \frac{\text{sgn}(t)}{\sqrt{2}}\cdot \lim_{t\rightarrow 0}(g(t)-g(0))\end{align*} Since the limit must be equal to $0$ it must holt that $\displaystyle{\lim_{t\rightarrow 0}g(t)=g(0)}$, which means that $g$ is continuous at $0$.

Is everything correct?

All correct (for evaluating the limit on the specified path). (Nod)

So we have proven that if $f$ is differentiable at $0$, that then $g$ must be continuous at $0$.
That leaves the converse.
How about switching to polar coordinates to prove the converse? (Wondering)
 
  • #7
I like Serena said:
It's given that $g$ is a function $\mathbb R \to \mathbb R$.
That means that $g$ is defined for every real number isn't it? (Wondering)

Ahh ok! (Nerd)
I like Serena said:
So we have proven that if $f$ is differentiable at $0$, that then $g$ must be continuous at $0$.
That leaves the converse.
How about switching to polar coordinates to prove the converse? (Wondering)

For the converse we suppose that $g$ is continuous at $x=0$ and we want to show that $f$ is differentiable at the origin, so we want to show that the three conditions of the first thread are satisfied, right?

Do you mean to use the polar coordinates at the third condition, at the limit?

(Wondering)
 
  • #8
mathmari said:
For the converse we suppose that $g$ is continuous at $x=0$ and we want to show that $f$ is differentiable at the origin, so we want to show that the three conditions of the first thread are satisfied, right?

Do you mean to use the polar coordinates at the third condition, at the limit?

Yes. (Thinking)
 
  • #9
I like Serena said:
Yes. (Thinking)

  • How can we check if $f_x(0,0)$ exists?
  • $f_y(0,0)$ exists

    We have that $f_y(x,y)=g(x)$. So $f_y(0,0)=g(0)$. Since $g(0)$ exists, it follows that $f_y(0,0)$ exists.
  • \begin{equation*} \\ \end{equation*}
    \begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{yg(x)-0\cdot g(0)-f_x(0,0)\cdot x-g(0)\cdot y}{\|(x,y)\|} \\ & = \lim_{(x,y)\rightarrow (0,0)}\frac{yg(x)-f_x(0,0)\cdot x-g(0)\cdot y}{\sqrt{x^2+y^2}} \\ & = \lim_{(x,y)\rightarrow (0,0)}\frac{y[g(x)-g(0)]-f_x(0,0)\cdot x}{\sqrt{x^2+y^2}}\end{align*} Using polar coordinates we have $x=r\cos\theta$ and $y=r\sin\theta$. Then we consider the limit that $r$ goes to $0$, don't we? (Wondering)
 
  • #10
mathmari said:
  • How can we check if $f_x(0,0)$ exists?

By treating $f_x(0,0)$ as an unknown in condition 3 and solving for it, as we did before. (Thinking)

mathmari said:
  • $f_y(0,0)$ exists

    We have that $f_y(x,y)=g(x)$. So $f_y(0,0)=g(0)$. Since $g(0)$ exists, it follows that $f_y(0,0)$ exists.

Indeed.
As a possible alternative, we could also solve $f_y(0,0)$ from condition 3.

mathmari said:
  • \begin{equation*} \\ \end{equation*}
    \begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{yg(x)-0\cdot g(0)-f_x(0,0)\cdot x-g(0)\cdot y}{\|(x,y)\|} \\ & = \lim_{(x,y)\rightarrow (0,0)}\frac{yg(x)-f_x(0,0)\cdot x-g(0)\cdot y}{\sqrt{x^2+y^2}} \\ & = \lim_{(x,y)\rightarrow (0,0)}\frac{y[g(x)-g(0)]-f_x(0,0)\cdot x}{\sqrt{x^2+y^2}}\end{align*} Using polar coordinates we have $x=r\cos\theta$ and $y=r\sin\theta$. Then we consider the limit that $r$ goes to $0$, don't we?

Yes. (Thinking)
 
  • #11
I like Serena said:
By treating $f_x(0,0)$ as an unknown in condition 3 and solving for it, as we did before. (Thinking)

But before why assumed that the condition 3 holds, i.e. that the limit is equal to 0. What do we have to do here? (Wondering)
I like Serena said:
Indeed.
As a possible alternative, we could also solve $f_y(0,0)$ from condition 3.

Ah ok!
 
  • #12
mathmari said:
But before why assumed that the condition 3 holds, i.e. that the limit is equal to 0. What do we have to do here?

Condition 3 is the definition of the derivative for multi-variable functions.
$f_x(0,0)$ and $f_y(0,0)$ are defined as the solutions of condition 3 - if we can find them.
If we can find them (they exist), it means that the function is differentiable.
And if we cannot find them, it means that the function is not differentiable.

Note that if we evaluate the limit following e.g. the path $(x,t)=(0,t)$, the limit in condition 3 collapses to the usual definition of the partial derivative. (Thinking)
 
  • #13
I haven't really understood that. (Thinking)
I like Serena said:
$f_x(0,0)$ and $f_y(0,0)$ are defined as the solutions of condition 3 - if we can find them.

Does does this mean? (Wondering)
I like Serena said:
Note that if we evaluate the limit following e.g. the path $(x,t)=(0,t)$, the limit in condition 3 collapses to the usual definition of the partial derivative. (Thinking)

Why? (Wondering)
 
  • #14
We already know that whatever g is, the dervative at (0,0) must be (0, g(0)), don't we?
That is, if f is differentiable at (0,0).

What is left is to verify that the limit of condition 3 exists, isn't it? (Wondering)
And we can use that g is continuous at 0.
 
  • #15
I like Serena said:
We already know that whatever g is, the dervative at (0,0) must be (0, g(0)), don't we?
That is, if f is differentiable at (0,0).

What is left is to verify that the limit of condition 3 exists, isn't it? (Wondering)
And we can use that g is continuous at 0.

\begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-0\cdot g(0)-f_x(0,0)\cdot x-g(0)\cdot y}{\|(x,y)\|} \\ & = \lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot [g(x)-g(0)]-f_x(0,0)\cdot x}{\sqrt{x^2+y^2}} \\ & = \lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot [g(x)-g(0)]}{\sqrt{x^2+y^2}}-\lim_{(x,y)\rightarrow (0,0)}\frac{f_x(0,0)\cdot x}{\sqrt{x^2+y^2}}\end{align*} The first limit is zero because $g$ is continuous at $0$, or can't we just say that?
What can we do for the second limit? I got stuck right now.

(Wondering)
 
  • #16
For the first part, no, we can't just say that it is zero.
We have to prove it for any path.
To do so, I suggest to switch to polar coordinates and see what happens. (Thinking)

For the second part, we know that the complete expressiion must be 0.
So if the first part is 0, then the second part must also be 0.
What does it take for it to be 0? (Wondering)
 
  • #17
I like Serena said:
For the first part, no, we can't just say that it is zero.
We have to prove it.
To do so, I suggest to switch to polar coordinates and see what happens. (Thinking)

Using polar coordinates $x=r\cos\theta, y=r\sin\theta$ we get: \begin{align*}\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot [g(x)-g(0)]}{\sqrt{x^2+y^2}}&=\lim_{r\rightarrow 0}\frac{r\sin\theta\cdot [g(r\cos\theta)-g(0)]}{r}=\lim_{r\rightarrow 0}\left (\sin\theta\cdot [g(r\cos\theta)-g(0)]\right )\\ &=\sin\theta\cdot\lim_{r\rightarrow 0} [g(r\cos\theta)-g(0)]\end{align*} If $r\rightarrow 0$ then $w:r\cos\theta\rightarrow 0$. So $\lim_{r\rightarrow 0} [g(r\cos\theta)-g(0)]=\lim_{w\rightarrow 0} [g(w)-g(0)]=0$, since $g$ is continuous at $0$.

Is everything correct?

I like Serena said:
For the second part, we know that the complete expressiion must be 0.
So if the first part is 0, then the second part must also be 0.
What does it take for it to be 0? (Wondering)

We have that \begin{equation*}\lim_{(x,y)\rightarrow (0,0)}\frac{f_x(0,0)\cdot x}{\sqrt{x^2+y^2}}=f_x(0,0)\cdot \lim_{(x,y)\rightarrow (0,0)}\frac{x}{\sqrt{x^2+y^2}}\end{equation*} Using again polar coordinates we get \begin{equation*}f_x(0,0)\cdot \lim_{(x,y)\rightarrow (0,0)}\frac{x}{\sqrt{x^2+y^2}}=f_x(0,0)\cdot \lim_{r\rightarrow 0}\frac{r\cos\theta}{r}=f_x(0,0)\cdot \lim_{r\rightarrow 0}\cos\theta=f_x(0,0)\cdot\cos\theta\end{equation*} right? Does this mean that $f_x(0,0)$ must be equal to $0$ ? (Wondering)
 
  • #18
All correct.
And yes, $f_x(0,0)$ must be equal to $0$, meaning it exists, and condition 3 holds. (Nod)
 
  • #19
I like Serena said:
All correct.
And yes, $f_x(0,0)$ must be equal to $0$, meaning it exists, and condition 3 holds. (Nod)

So, at this direction we do the following:
We suppose that $g$ is continuous at $x=0$.
Then we have $f_y(x,y)=g(x)$. At $(0,0)$ we get $f_y(0,0)=g(0)$ which exists.

The third condition holds, i.e. the limit is equal to $0$ if and only if $f_x(0,0)$ is equal to $0$ (so it exists). That means either all condition hold, or just the second one.
Have I understood that correctly? (Wondering)
 
  • #20
mathmari said:
So, at this direction we do the following:
We suppose that $g$ is continuous at $x=0$.
Then we have $f_y(x,y)=g(x)$. At $(0,0)$ we get $f_y(0,0)=g(0)$ which exists.

The third condition holds, i.e. the limit is equal to $0$ if and only if $f_x(0,0)$ is equal to $0$ (so it exists). That means either all condition hold, or just the second one.
Have I understood that correctly? (Wondering)

Yes. (Nod)
And for differentiability it suffices if we can find a $f_x(0,0)$ and $f_y(0,0)$ such that all 3 conditions hold.
The 3rd condition effectively determines $f_x(0,0)$ and $f_y(0,0)$.

There is a theorem that says that $f_x(0,0)$ and $f_y(0,0)$ will be equal to the partial derivative functions at (0,0) if the partial derivative functions are well-defined on a neighborhood. This is what we used for $f_y(0,0)$, but what we couldn't use for $f_x(0,0)$. (Nerd)
 
  • #21
I like Serena said:
Yes. (Nod)
And for differentiability it suffices if we can find a $f_x(0,0)$ and $f_y(0,0)$ such that all 3 conditions hold.
The 3rd condition effectively determines $f_x(0,0)$ and $f_y(0,0)$.

There is a theorem that says that $f_x(0,0)$ and $f_y(0,0)$ will be equal to the partial derivative functions at (0,0) if the partial derivative functions are well-defined on a neighborhood. This is what we used for $f_y(0,0)$, but what we couldn't use for $f_x(0,0)$. (Nerd)

Ah ok! Thanks a lot! (Yes)
 

FAQ: Is Differentiability at the Origin Determined by Partial Derivatives?

1) What is the concept of differentiability in origin?

Differentiability in origin refers to the property of a mathematical function where it has a well-defined derivative at the origin, or where x=0. This means that the function is smooth and continuous at this point.

2) How is differentiability in origin determined?

Differentiability in origin can be determined by evaluating the derivative of a function at the origin. If the derivative exists and is finite, the function is said to be differentiable at the origin.

3) What is the significance of differentiability in origin?

Differentiability in origin is important in calculus and mathematical analysis, as it allows for the use of certain mathematical techniques and theorems. It also indicates that the function is well-behaved and has a smooth curve at the origin.

4) Can a function be differentiable at the origin but not continuous?

No, a function cannot be differentiable at the origin if it is not continuous at that point. This is because differentiability requires continuity, but the converse is not necessarily true.

5) How does differentiability in origin differ from differentiability in general?

Differentiability in origin is a specific case of differentiability, where the function is evaluated at the point x=0. In general, differentiability refers to the property of a function having a well-defined derivative at all points in its domain.

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