- #1

The Rahul

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Hii please help me to Solve problems diffrentiability of function with two variable with one method.

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In summary, The Rahul found a way to solve a problem that involves finding the derivative of a function at a particular point, but found an easier way to do it using x=y3.

- #1

The Rahul

- 5

- 0

Hii please help me to Solve problems diffrentiability of function with two variable with one method.

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- #2

tiny-tim

Science Advisor

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Hi The Rahul! Welcome to PF!

Show us one or two examples that you've had difficulty with, together with your attempts to solve them.

- #3

Evo

Staff Emeritus

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The Rahul said:Hii Tiny tim,thnx for ur warm welcome,

Now I continue my problem=

In this Ques.—f(x,y)={|xy|}^1/2 Not differentiable at (0,0)

I find out Both Partial derivative fx and fy and solve the ques.

But in that ques.== f(x,y)= xy /√(x^2+y^2)

in my book method is diffrent.

in that ques method is to find according to y=mx and x=y^3.

so Plz help me...:-D

- #4

The Rahul

- 5

- 0

Hii Tiny tim,thnx for ur warm welcome,

Now I continue my problem=

In this Ques.—f(x,y)={|xy|}

I find out Both Partial derivative f

But in that ques.== f(x,y)= xy /√(x

in my book method is diffrent.

in that ques method is to find according to y=mx and x=y

so Plz help me...:-D

- #5

tiny-tim

Science Advisor

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To prove that a function f(x,y) is

we only need to find

If f(x,y) = xy /√(x^{2}+y^{2}) = xy/r,

then the derivative along any straight line*does* exist at (0,0), so we *can't* stop there, we need to check* other* ways of approaching the origin …

in this case, the easiest curve to check is x=y^{3} (or y=x^{3})

(x=y^{2} is awkward, because it gives you awkward square-roots

then the derivative along any straight line

in this case, the easiest curve to check is x=y

(x=y

btw,

- #6

The Rahul

- 5

- 0

tiny-tim said:

To prove that a function f(x,y) isnotdifferentiable at (0,0),

we only need to findonecurve along which it is not differentiable (and then we can stop).

If f(x,y) = xy /√(x^{2}+y^{2}) = xy/r,

then the derivative along any straight linedoesexist at (0,0), so wecan'tstop there, we need to checkotherways of approaching the origin …

in this case, the easiest curve to check is x=y^{3}(or y=x^{3})

(x=y^{2}is awkward, because it gives you awkward square-roots

btw,since you've woken the mentors, please note that txt-english (eg "please", "ur") isagainst the forum rules!

Thank you for your precious reply.It works.

- #7

HallsofIvy

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Differentiability for a function with two variables means that the function has a well-defined tangent plane at every point in its domain. This means that the function is smooth and has no sharp turns or corners.

Partial derivatives measure the rate of change of a function with respect to one variable while holding all other variables constant. Total derivatives, on the other hand, measure the overall rate of change of a function with respect to all its variables.

A function with two variables is differentiable at a specific point if all its partial derivatives exist and are continuous at that point. This means that the function is smooth and has a well-defined tangent plane at that point.

A function with two variables can be differentiable without being continuous, but it cannot be continuous without being differentiable. This means that a function must be continuous in order to be differentiable, but being differentiable does not necessarily guarantee continuity.

Differentiability can be used to determine the local behavior of a function with two variables, such as the existence of maxima and minima. It can also be used to approximate the value of a function at a specific point using its tangent plane.

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