Proving Differentiability of f(x,y) = x(|y|^(1/2)) at (0,0)

In summary: So the partial derivative ∂f/∂y is always zero at the origin for every function, no matter what the directional derivative might be.In summary, you can differentiate f at the origin using the partial derivative with respect to y, and the directional derivative will always be zero.
  • #1
samithie
7
0
f: R2 to R1 given by f(x,y) = x(|y|^(1/2))
show differentiable at (0,0)

so I'm using the definition lim |h| ->0 (f((0,0) + 9(h1,h2)) - f(0,0) - Df(0,0) (h1,h2)) / |h|

so first for the jacobian for f, when I'm doing the partial with respect to y, do I have to break this into the case y>0 and y<0 and show its differentiable in both cases (and maybe also have to do the same for when h2 >0 or <0) or can I use do it in one step by rewriting and by differentiating (y^2)^1/4 and. I did that and the Df(0,0) just goes away and then limit doesn't go to 0 it seems. Any help is appreciated thanks!
 
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  • #2
samithie said:
f: R2 to R1 given by f(x,y) = x(|y|^(1/2))
show differentiable at (0,0)

so I'm using the definition lim |h| ->0 (f((0,0) + 9(h1,h2)) - f(0,0) - Df(0,0) (h1,h2)) / |h|

so first for the jacobian for f, when I'm doing the partial with respect to y, do I have to break this into the case y>0 and y<0 and show its differentiable in both cases (and maybe also have to do the same for when h2 >0 or <0) or can I use do it in one step by rewriting and by differentiating (y^2)^1/4 and. I did that and the Df(0,0) just goes away and then limit doesn't go to 0 it seems. Any help is appreciated thanks!
The title of this thread is misleading. This thread has to do differentiation, rather than integration.

It looks like you have some typo's in your definition (of whatever, you don't say),
lim |h| ->0 (f((0,0) + 9(h1,h2)) - f(0,0) - Df(0,0) (h1,h2)) / |h|​

Clearly, the partial derivative, ∂f(x,y)/∂y, does not exist for y=0, except perhaps when x=0 and y=0 . Of course, that's at least part the matter at hand ...

I'm no expert on this, but I suggest looking at the directional derivative of f(x,y) at (0,0), for arbitrary direction.
 
  • #3
yes sorry about the typos. I think the directional derivatives is not the issue I'm just wondering about if I need to do this piecewise.
 
  • #4
samithie said:
yes sorry about the typos. I think the directional derivatives is not the issue I'm just wondering about if I need to do this piecewise.
√(|y|) is equivalent to (y2)1/4
 

1. What is the definition of absolute value?

The absolute value of a number is its distance from 0 on a number line. It is always a positive value.

2. How is the absolute value symbol represented in mathematical notation?

The absolute value symbol is represented by a pair of vertical lines surrounding the number. For example, |x| represents the absolute value of x.

3. What is the difference between integrating a function with absolute value and without absolute value?

Integrating a function with absolute value involves breaking the integral into multiple parts, one for each interval where the function changes sign. This is because the absolute value function is not continuous and has a "kink" at the origin. Integrating a function without absolute value does not require this step.

4. Can the integration of absolute value be applied to any type of function?

Yes, the integration of absolute value can be applied to any type of function as long as it is defined on a closed interval and satisfies certain continuity conditions.

5. What are some real-world applications of integrating absolute value?

The integration of absolute value is commonly used in physics and engineering to calculate the displacement of an object, as well as in economics to calculate the total cost of a product with changing prices. It is also used in signal processing to analyze and filter digital signals.

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