# Thread: Proof that f(x,y) is differentiable and determine the derivate

1. ## Proof that f(x,y) is differentiable and determine the derivate

Hey guys!

I am supposed to show that the function f(x,y) = x^2 + y^3 is differentiable and to determine the derivate.

Do i have to show this with the differential quotient? I know about partial derivates but how to determine the "entire derivate" f'(x,y)?

I know lots of questions and no solution. Sorry for that, but please give me a clue, some keywords at least.

Thanks!

Inf

2. Originally Posted by Inf
Hey guys!

I am supposed to show that the function f(x,y) = x^2 + y^3 is differentiable and to determine the derivate.

Do i have to show this with the differential quotient? I know about partial derivates but how to determine the "entire derivate" f'(x,y)?

I know lots of questions and no solution. Sorry for that, but please give me a clue, some keywords at least.

Thanks!

Inf
Dear Inf,

For a multivariable function(a function with two or more variables) the derivative is not defined. Only the partial derivative is defined. So there is no such thing as the "entire derivative."

3. Dear Sudharaka,

So, these are my partial derivates:
df(x,y)/dx = 2x
df(x,y)/dy = 3y^2

Did i proof now that f is differentiable?

Inf

4. Originally Posted by Inf
Dear Sudharaka,

So, these are my partial derivates:
df(x,y)/dx = 2x
df(x,y)/dy = 3y^2

Did i proof now that f is differentiable?

Inf
If you need to do this by first principles then

since this is a function from

$f:\mathbb{R}^2 \to \mathbb{R}$

The total derivative will be a 1 by 2 matrix

The definition is

$\lim_{\mathbf{h} \to \mathbf{0}} \bigg{|}\bigg{|}\frac{f(\mathbf{x}+\mathbf{h})- f(\mathbf{x})- \mathbf{J} \mathbf{h}}{\mathbf{h}} \bigg{|}\bigg{|} =0$

Where

$\mathbf{x} = \begin{bmatrix}x \\ y \end{bmatrix} \quad \mathbf{h} = \begin{bmatrix}h_1 \\ h_2 \end{bmatrix}$

and J is the Jacobian matrix (in this case the gradient of f)

$\mathbf{J} = \begin{bmatrix}2x & 3y^2 \end{bmatrix}$

This will prove that the Jacobian is the derivative.

5. Hey TheEmptySet,

thank you! After doing some "calculation", i get
$\lim\limits_{h\rightarrow 0}{}\frac{1}{||h||}\cdot(h^3_2 + 3h^2_{2}y + h^2_1) = 0$

||.|| is a norm, right? which one applies here? euclidean one?

$||h|| = \sqrt{h^2_1 + h^2_2}$

and if it is this one how to move on?

Inf

6. Note that

$\bigg|\bigg| \frac{h_1^2+3yh_2^2+h_2^3}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| \le \bigg|\bigg|\frac{h_1^2}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| +\bigg|\bigg| \frac{3yh_2^2}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| +\bigg|\bigg| \frac{h_2^3}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg|$

$\le |h_1|+3|y||h_2| +|h_2^2|$

7. Ok, thanks!

Bunch of questions

Why are you using a norm on a real number?
Why are you using inequalities?
Where is the limit?

Cheers,
Inf

8. Originally Posted by Inf
Ok, thanks!

Bunch of questions

Why are you using a norm on a real number?
Why are you using inequalities?
Where is the limit?

Cheers,
Inf
The norm of a scalar is just its absolute value

The limit itself is hard to evaluate directly but if you use the inequalities above you can use the squeeze theorem

This gives

$0 \le \bigg|\bigg| \frac{h_1^2+3yh_2^2+h_2^3}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| \le \bigg|\bigg|\frac{h_1^2}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| +\bigg|\bigg| \frac{3yh_2^2}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| +\bigg|\bigg| \frac{h_2^3}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| \le |h_1|+3|y||h_2| +|h_2^2|$

Now if you take the limit as h goes to zero you get

$0 \le \bigg|\bigg| \frac{h_1^2+3yh_2^2+h_2^3}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| \le 0$

So the limit must be zero.

9. Thank you, now i get it!

Why
$|h_1|+3|y||h_2| +|h_2^2|$
and not
$|h_1^2|+3|y||h_2^2| +|h_2^3|$
?

"h goes to zero" means both $h_1$ and $h_2$ go to zero, right?

Inf

10. Originally Posted by Inf
Thank you, now i get it!

Why
$|h_1|+3|y||h_2| +|h_2^2|$
and not
$|h_1^2|+3|y||h_2^2| +|h_2^3|$
?

"h goes to zero" means both $h_1$ and $h_2$ go to zero, right?

Inf
I will do the first one the other two are similar. The problem is that the denominator is also going to zero.

$\bigg|\bigg|\frac{h_1^2}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg|=|h_1|\bigg|\bigg|\frac{h_1}{\sqrt{h_1 ^2+h_2^2}} \bigg|\bigg|$

Note that

$h_1^2 \le h_1^2+h_2^2 \implies |h_1| < \sqrt{h_1^2+h_2^2} \iff \frac{|h_1|}{\sqrt{h_1^2+h_2^2}} \le 1$

Now putting this into the above gives

$|h_1|\bigg|\bigg|\frac{h_1}{\sqrt{h_1^2+h_2^2}} \bigg|\bigg| \le |h_1|\cdot 1$

Yes you need to let both

$h_1 \to 0 \text{ and } h_2 \to 0$

11. Thank you, now i get it!

Why
$|h_1|+3|y||h_2| +|h_2^2|$
and not
$|h_1^2|+3|y||h_2^2| +|h_2^3|$
?

"h goes to zero" means both $h_1$ and $h_2$ go to zero, right?

Inf

12. Originally Posted by Sudharaka
Dear Inf,

For a multivariable function(a function with two or more variables) the derivative is not defined. Only the partial derivative is defined. So there is no such thing as the "entire derivative."
That's simply not true- but of course, you have to generalize the notion of "derivative". If f is a function from $R^n$ to $R^m$ then the derivative of f, at point p, is the linear function from $R^n$ to $R^m$, that best approximates f in a neighborhood of p. Specifically, we say that such a function is differentiable at p if and only if there exist a linear function L and a function $\epsilon(x)$, from $R^n$ to $R^m$, such that
$f(x)= f(p)+ L(x- p)+ \epsilon(x)$
$\lim_{x\to p}\frac{\epsilon(x)}{||x- p||}= 0$

Notice the the derivative is defined as a linear transformation. In the case of $R^1$ to $R^1$, it would be the function $y= \frac{df}{dx}(p)x$ where $\frac{df}{dx}(p)$ is the "usual" derivative at p. In the case of $R^2$ to $R^1$ the linear transformation is maps $(p_x, p_y)$ to $\frac{\partial f}{\partial x}p_x+ \frac{\partial f}{\partial y}p_y$. In other words it is the dot product of $\nabla f$ and p. That's why most texts treat $\nabla f$ as being "the" derivative.

13. Ok, now i finally understand! thank you guys!
Sorry for the double post above..