Find partial derivatives of $f(x,y)= sqrt xy$ at the point $(0,0)$
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I'm trying to find partial derivatives $f_x$ and $f_y$ at $(0,0)$, assuming they exist (although I believe they don't), of the function $f(x,y)= sqrt xy$
when I use the chain rule, I get partial derivatives that would result in $0over 0$
e.g. $yover 2sqrtleft(xyright)$ and $xover 2sqrtleft(xyright)$
I assume I need to use the first principle approach to find a valid answer.
So far I have, for $f_x$ using the point $(0,0)$
$f_x =lim_hto 0sqrt left((x+h)yright) - sqrtleft(xyright)overh$
$=lim_hto 0sqrt(left(0+h)0right) - sqrtleft(0right)overh$
which is the same as
$=lim_hto 0sqrt0-sqrt0overh$
which equals zero
Is this a valid way to find that the partial derivative for $f_x = 0$? Is it even zero?
I have followed the same method for $f_y$
Am I on the correct track here or am I barking up the wrong tree?
calculus derivatives partial-derivative
add a comment |Â
up vote
0
down vote
favorite
I'm trying to find partial derivatives $f_x$ and $f_y$ at $(0,0)$, assuming they exist (although I believe they don't), of the function $f(x,y)= sqrt xy$
when I use the chain rule, I get partial derivatives that would result in $0over 0$
e.g. $yover 2sqrtleft(xyright)$ and $xover 2sqrtleft(xyright)$
I assume I need to use the first principle approach to find a valid answer.
So far I have, for $f_x$ using the point $(0,0)$
$f_x =lim_hto 0sqrt left((x+h)yright) - sqrtleft(xyright)overh$
$=lim_hto 0sqrt(left(0+h)0right) - sqrtleft(0right)overh$
which is the same as
$=lim_hto 0sqrt0-sqrt0overh$
which equals zero
Is this a valid way to find that the partial derivative for $f_x = 0$? Is it even zero?
I have followed the same method for $f_y$
Am I on the correct track here or am I barking up the wrong tree?
calculus derivatives partial-derivative
The partial derivatives do exist, but this $f$ is not differentiable at the origin, which is a stronger condition.
â amd
Aug 15 at 19:48
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to find partial derivatives $f_x$ and $f_y$ at $(0,0)$, assuming they exist (although I believe they don't), of the function $f(x,y)= sqrt xy$
when I use the chain rule, I get partial derivatives that would result in $0over 0$
e.g. $yover 2sqrtleft(xyright)$ and $xover 2sqrtleft(xyright)$
I assume I need to use the first principle approach to find a valid answer.
So far I have, for $f_x$ using the point $(0,0)$
$f_x =lim_hto 0sqrt left((x+h)yright) - sqrtleft(xyright)overh$
$=lim_hto 0sqrt(left(0+h)0right) - sqrtleft(0right)overh$
which is the same as
$=lim_hto 0sqrt0-sqrt0overh$
which equals zero
Is this a valid way to find that the partial derivative for $f_x = 0$? Is it even zero?
I have followed the same method for $f_y$
Am I on the correct track here or am I barking up the wrong tree?
calculus derivatives partial-derivative
I'm trying to find partial derivatives $f_x$ and $f_y$ at $(0,0)$, assuming they exist (although I believe they don't), of the function $f(x,y)= sqrt xy$
when I use the chain rule, I get partial derivatives that would result in $0over 0$
e.g. $yover 2sqrtleft(xyright)$ and $xover 2sqrtleft(xyright)$
I assume I need to use the first principle approach to find a valid answer.
So far I have, for $f_x$ using the point $(0,0)$
$f_x =lim_hto 0sqrt left((x+h)yright) - sqrtleft(xyright)overh$
$=lim_hto 0sqrt(left(0+h)0right) - sqrtleft(0right)overh$
which is the same as
$=lim_hto 0sqrt0-sqrt0overh$
which equals zero
Is this a valid way to find that the partial derivative for $f_x = 0$? Is it even zero?
I have followed the same method for $f_y$
Am I on the correct track here or am I barking up the wrong tree?
calculus derivatives partial-derivative
edited Aug 15 at 11:02
asked Aug 15 at 10:39
Sami Burton
404
404
The partial derivatives do exist, but this $f$ is not differentiable at the origin, which is a stronger condition.
â amd
Aug 15 at 19:48
add a comment |Â
The partial derivatives do exist, but this $f$ is not differentiable at the origin, which is a stronger condition.
â amd
Aug 15 at 19:48
The partial derivatives do exist, but this $f$ is not differentiable at the origin, which is a stronger condition.
â amd
Aug 15 at 19:48
The partial derivatives do exist, but this $f$ is not differentiable at the origin, which is a stronger condition.
â amd
Aug 15 at 19:48
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
The partial derivatives of $f$ at the origin are both $0$, and you calculated $f_x(0,0)$ correctly. $f_y$ follows in more or less the same manner.
I have a habit of overthinking! I know the that the answer is zero, but was just unsure if my working was sufficient. Thank You.
â Sami Burton
Aug 15 at 10:49
Would the use of the squeeze theorem be a better answer?
â Sami Burton
Aug 15 at 10:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The partial derivatives of $f$ at the origin are both $0$, and you calculated $f_x(0,0)$ correctly. $f_y$ follows in more or less the same manner.
I have a habit of overthinking! I know the that the answer is zero, but was just unsure if my working was sufficient. Thank You.
â Sami Burton
Aug 15 at 10:49
Would the use of the squeeze theorem be a better answer?
â Sami Burton
Aug 15 at 10:57
add a comment |Â
up vote
0
down vote
accepted
The partial derivatives of $f$ at the origin are both $0$, and you calculated $f_x(0,0)$ correctly. $f_y$ follows in more or less the same manner.
I have a habit of overthinking! I know the that the answer is zero, but was just unsure if my working was sufficient. Thank You.
â Sami Burton
Aug 15 at 10:49
Would the use of the squeeze theorem be a better answer?
â Sami Burton
Aug 15 at 10:57
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The partial derivatives of $f$ at the origin are both $0$, and you calculated $f_x(0,0)$ correctly. $f_y$ follows in more or less the same manner.
The partial derivatives of $f$ at the origin are both $0$, and you calculated $f_x(0,0)$ correctly. $f_y$ follows in more or less the same manner.
answered Aug 15 at 10:43
Arthur
99.8k793175
99.8k793175
I have a habit of overthinking! I know the that the answer is zero, but was just unsure if my working was sufficient. Thank You.
â Sami Burton
Aug 15 at 10:49
Would the use of the squeeze theorem be a better answer?
â Sami Burton
Aug 15 at 10:57
add a comment |Â
I have a habit of overthinking! I know the that the answer is zero, but was just unsure if my working was sufficient. Thank You.
â Sami Burton
Aug 15 at 10:49
Would the use of the squeeze theorem be a better answer?
â Sami Burton
Aug 15 at 10:57
I have a habit of overthinking! I know the that the answer is zero, but was just unsure if my working was sufficient. Thank You.
â Sami Burton
Aug 15 at 10:49
I have a habit of overthinking! I know the that the answer is zero, but was just unsure if my working was sufficient. Thank You.
â Sami Burton
Aug 15 at 10:49
Would the use of the squeeze theorem be a better answer?
â Sami Burton
Aug 15 at 10:57
Would the use of the squeeze theorem be a better answer?
â Sami Burton
Aug 15 at 10:57
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2883460%2ffind-partial-derivatives-of-fx-y-sqrt-xy-at-the-point-0-0%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
The partial derivatives do exist, but this $f$ is not differentiable at the origin, which is a stronger condition.
â amd
Aug 15 at 19:48