Question

limit definition of partial derivatives of. Tedious Algebra !!!

Answer 1

\[gx(X,Y)= (x^3-x^2y)/x^2+y^2\]

Answer 2

\[[(x+h)^3-(x+h)^2y]/(x+h)^2+y^2 - [x^3-x^2y]/x^2+y^2\]

Answer 3

very messy.

Answer 4

@Abhisar

Answer 5

@IrishBoy123

Answer 6

I'm guessing you're given this g(x,y) function

\[\Large g(x,y) = \frac{x^3-x^2y}{x^2+y^2}\]

right? or no?

Answer 7

\(g_x\).... ?

Answer 8

subscript that denotes partial derivative wrt x

Answer 9

Anyway, won't interrupt...

Answer 10

yes please

Answer 11

@SolomonZelman nobody is answering the question

Answer 12

so you are not interupting

Answer 13

you are doing differentiation, SAME THING!
Regard everything besides "x" as constant

Answer 14

and x is the variable with respect to which you differentiate

Answer 15

u sure you want to do the first principles??

Answer 16

https://www.youtube.com/watch?v=powyITyDfgI
That should probably help you

Answer 17

@SolomonZelman we are to use limit definition of partial derivatives

Answer 18

\(\color{#000000 }{ \displaystyle \lim_{h\to 0}\frac{\dfrac{(x+h)^3-(x+h)^2y}{(x+h)^2+(y+h)^2}-\dfrac{x^3-x^2y}{x^2+y^2} }{h} }\)

Answer 19

your instructor is ruthless ...

we need algebra here ...

Answer 20

did I write it correctly, is that fraction on the left top your f ?

Answer 21

Looks good.

Answer 22

oh, k, you answered my Q.

Answer 23

yea that's the function

Answer 24

\(\color{#000000 }{ \displaystyle \lim_{h\to 0}\frac{\dfrac{\color{blue}{(x+h)^3-(x+h)^2y}}{\color{red}{(x+h)^2+(y+h)^2}}-\dfrac{x^3-x^2y}{x^2+y^2} }{h} }\)

expanding the blue and the red, is where I would start...

Answer 25

HELLO

Answer 26

WHY DO YOU HAVE (Y+H)^2

Answer 27

ohhh

Answer 28

we are finding gx

Answer 29

\(\color{#000000 }{ \displaystyle \lim_{h\to 0}\frac{\dfrac{(x+h)^3-(x+h)^2y}{(x+h)^2+y^2}-\dfrac{x^3-x^2y}{x^2+y^2} }{h} }\)

Answer 30

yes, I am sorry ...

Answer 31

yap.

Answer 32

\(\color{#000000 }{ \tt \displaystyle \lim_{h\to 0}\frac{\dfrac{(x+h)^3-(x+h)^2y}{(x+h)^2+y^2}-\dfrac{x^3-x^2y}{x^2+y^2} }{h} }\)

\(\color{#000000 }{ \tt \displaystyle \lim_{h\to 0}\frac{\dfrac{\left[(x+h)^3-(x+h)^2y\right]\left[x^2+y^2\right]-\left[(x+h)^2+y^2\right]\left[x^3-x^2y\right]}{\left[(x+h)^2+y^2\right]\left[x^2+y^2\right]} }{h} }\)

\(\color{#000000 }{ \tt \displaystyle \lim_{h\to 0}\frac{\dfrac{\left[(x+h)^3-(x+h)^2y\right]\left[x^2+y^2-x^3+x^2y\right]}{\left[(x+h)^2+y^2\right]\left[x^2+y^2\right]} }{h} }\)

\(\color{#000000 }{ \tt \displaystyle \lim_{h\to 0}\frac{\dfrac{\left[(x+h)^3-(x+h)^2y\right]\left[x^2+y^2-x^3+x^2y\right]}{\left[(x+h)^2+y^2\right]\left[x^2+y^2\right]} }{h} }\)

this is very long to type....
but the expension on top and bottom should follow and cancel the h's

Answer 33

I will do on paper ...

Answer 34

expanding is like really ridiculous, but using the limit definition i just see no way :(

Answer 35

I am sure you know how to expand, all it is - just pain, so you can expand top and bottom via wolfram, if you like...

Answer 36

yea am on it

Answer 37

you expanded it?
you are a hero then:)

Answer 38

lol I am trying to expand it

Answer 39

I will differentiate for confirmation, using normal derivative ...

Answer 40

ok.

Answer 41

one more thing before you go, actually its part of my question. whats the limit of that function. d/dx?

Answer 42

I am guessing we are to use polar co-ordinates

Answer 43

\(\color{#000000 }{ \displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}}\)

Answer 44

that ?

Answer 45

Using the definition of continuity at a point determine whether gx
is continuous at (0,0) . Show all work.

Answer 46

yes

Answer 47

I am still expanding lol

Answer 48

\[\Large g(x,y) = \frac{x^3-x^2y}{x^2+y^2}\]

\[\Large g(x+h,y) = \frac{(x+h)^3-(x+h)^2y}{(x+h)^2+y^2}\]



\[\large g(x+h,y) - g(x,y)= \frac{(x+h)^3-(x+h)^2y}{(x+h)^2+y^2}-\frac{x^3-x^2y}{x^2+y^2}\]



Let


A = (x+h)^3-(x+h)^2y
B = (x+h)^2+y^2
C = x^3-x^2y
D = x^2+y^2


The majority of the work (in my opinion) lies on computing g(x+h,y) - g(x,y) (expanding and simplifying) so let's do that using A,B,C,D defined above


\[\large g(x+h,y) - g(x,y) = \frac{A}{B} - \frac{C}{D}\]
\[\large g(x+h,y) - g(x,y) = \frac{AD-BC}{BD}\]


A*D = ((x+h)^3-(x+h)^2y)*(x^2+y^2)
A*D = (x^3+3x^2h+3xh^2+h^3-x^2y-2xhy-h^2y)*(x^2+y^2)
A*D = (x^2+y^2)(x^3+3x^2h+3xh^2+h^3-x^2y-2xhy-h^2y)
A*D = x^2*(x^3+3x^2h+3xh^2+h^3-x^2y-2xhy-h^2y)+y^2*(x^3+3x^2h+3xh^2+h^3-x^2y-2xhy-h^2y)
A*D = x^5+3*x^4*h+3*x^2*xh^2+x^2*h^3-x^4*y-2*x^2*xhy-x^2*h^2*y+y^2*x^3+3*y^2*x^2*h+3*y^2*xh^2+y^2*h^3-y^3*x^2-2*y^2*xhy-y^3*h^2
A*D = x^5+3*x^4*h+3*x^2*xh^2+x^2*h^3-x^4*y-2*x^2*xhy-x^2*h^2*y+y^2*x^3+3*y^2*x^2*h+3*y^2*xh^2+y^2*h^3-y^3*x^2-2*y^2*xhy-y^3*h^2


B*C = ((x+h)^2+y^2)*(x^3-x^2y)
B*C = (x^2+2xh+h^2+y^2)*(x^3-x^2y)
B*C = (x^3-x^2y)*(x^2+2xh+h^2+y^2)
B*C = x^3(x^2+2xh+h^2+y^2)-x^2y(x^2+2xh+h^2+y^2)
B*C = x^5+2x^4h+x^3h^2+x^3y^2-x^4y-2x^3hy-x^2h^2y-x^2y^3



Now compute A*D - B*C


(x^5+3*x^4*h+3*x^2*xh^2+x^2*h^3-x^4*y-2*x^2*xhy-x^2*h^2*y+y^2*x^3+3*y^2*x^2*h+3*y^2*xh^2+y^2*h^3-y^3*x^2-2*y^2*xhy-y^3*h^2)-(x^5+2x^4h+x^3h^2+x^3y^2-x^4y-2x^3hy-x^2h^2y-x^2y^3)

x^4*h+2*x^3*h^2+x^2*h^3+3*y^2*x^2*h+3*y^2*x*h^2+y^2*h^3-2*y^3*x*h-y^3*h^2

h(x^4+2*x^3*h+x^2*h^2+3*x^2*y^2+3*x*y^2*h+y^2*h^2-2x*y^3-y^3*h)


The factored out 'h' will cancel with the h in the very bottom denominator leaving you with


x^4+2*x^3*h+x^2*h^2+3*x^2*y^2+3*x*y^2*h+y^2*h^2-2x*y^3-y^3*h


Now apply the limit as h goes to 0 to get

x^4+3*x^2*y^2-2x*y^3

This is the numerator. The denominator will be B*D = ((x+h)^2+y^2)*(x^2+y^2). When h goes to 0, you will get (x^2+y^2)^2


So overall, the partial with respect to x is

\[\Large g_x(x,y) = \frac{x^4+3x^2y^2-2xy^3}{(x^2+y^2)^2}\]

Answer 49

and every limit defintion problem is like this ....

Answer 50

I am speechless. WOW. @jim_thompson5910

Answer 51

:) That was just awesome :)

Answer 52

Anyway, ... using the definition of continuity;

if we have
\(\color{#000000 }{ \displaystyle f(x,y)=\frac{x^3-x^2y}{x^2+y^2}}\)

then, for it to be continuous at (0,0)
then, the following limit exists.

\(\color{#000000 }{ \displaystyle \lim_{(x,y)\to(0,0)}\frac{x^3-x^2y}{x^2+y^2}}\)

Answer 53

NO NO NO NO NO

Answer 54

But, f(0,0) has to exist also

Answer 55

THE DERIVATIVE OF THE FUNCTION.

Answer 56

Using the definition of continuity at a point determine whether gx
is continuous at (0,0) . Show all work.

Answer 57

I was thinking that the function is continuous if

\(\color{#000000 }{ \displaystyle \lim_{(x,y)\to(a,b)}f(x,y)=(a,b)}\)

\(\color{#000000 }{ \displaystyle f(a,b)}\) exists

Answer 58

yes.

Answer 59

we use polar coordinates right?

Answer 60

exactly as @SolomonZelman is saying

it turns out that the limit doesn't exist. The way to show this is to follow two different paths approaching (0,0) and show that g_x is different

Answer 61

f(0,0) does not exist

Answer 62

Ok I WILL DO JUST THAT.

Answer 63

yupp

Answer 64

no need differentiating once you see f(a,b) doesn't exist... then you know it is discountinues at (a,b)...

Answer 65

@jim_thompson5910 I approach from the X axis and the limit is 1

Answer 66

both conditions (the once I mentioned for continuity) have to be satisfied.

Answer 67

If you follow the path along the x axis, then y = 0



\[\Large g_x(x,y) = \frac{x^4+3x^2y^2-2xy^3}{(x^2+y^2)^2}\]

\[\Large g_x(x,0) = \frac{x^4+3x^2(0)^2-2x(0)^3}{(x^2+(0)^2)^2}\]
\[\Large g_x(x,0) = \frac{x^4}{(x^2)^2}\]
\[\Large g_x(x,0) = \frac{x^4}{x^4}\]
\[\Large g_x(x,0) = 1\]

Apply the limit as x ---> 0 and you'll get 1 as the limiting value. Keep this value (1) in mind.


--------------------------------------------------------


Now follow the path along the line y = x


\[\Large g_x(x,y) = \frac{x^4+3x^2y^2-2xy^3}{(x^2+y^2)^2}\]

\[\Large g_x(x,x) = \frac{x^4+3x^2x^2-2x*x^3}{(x^2+x^2)^2}\]

\[\Large g_x(x,x) = \frac{x^4+3x^4-2x^4}{(2x^2)^2}\]

\[\Large g_x(x,x) = \frac{2x^4}{4x^4}\]

\[\Large g_x(x,x) = \frac{1}{2}\]

as you take the limit as x ---> 0, then the limiting value is 1/2 which is different from the previous limiting value of 1

Answer 68

@SolomonZelman has an easier approach, but it's always good to know how to do it using limits

Answer 69

yes, because most problems will have the function defined at f(a,b)

Answer 70

in fact 99.9999 % of them

Answer 71

anyways.... good luck :)

Answer 72

oh mehn thank you so much. who wants the medal @jim_thompson5910 and @SolomonZelman but i have one last problem related to this problem.

Answer 73

doesn't matter ....

Answer 74

Thank you so much. really appreciate.

Answer 75

anytime, for all that I know in math.

Answer 76

calculate gx(X Y) for all (xy) NOT EQUAL TO 0. simplify your answer. @SolomonZelman

Answer 77

the derivative with respect to x?
Haven't we already done that?

Answer 78

yes but for all (x,y) NOT EQUAL TO 0

Answer 79

\(\color{#000000 }{ \displaystyle g_x(x,y)=\frac{x^4+3x^2y^2-2xy^3}{(x^2+y^2)^2} }\)

Answer 80

please see question 2 @SolomonZelman

Answer 81

2a to be precisely

Answer 82

That is the derivative, I think ...
they don't mean what you meant

Answer 83

the function is peacewise, and has to parts