Find partial derivatives of $f(x,y)= sqrt xy$ at the point $(0,0)$

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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?







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  • The partial derivatives do exist, but this $f$ is not differentiable at the origin, which is a stronger condition.
    – amd
    Aug 15 at 19:48















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?







share|cite|improve this question






















  • The partial derivatives do exist, but this $f$ is not differentiable at the origin, which is a stronger condition.
    – amd
    Aug 15 at 19:48













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?







share|cite|improve this question














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?









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 15 at 11:02

























asked Aug 15 at 10:39









Sami Burton

404




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  • The partial derivatives do exist, but this $f$ is not differentiable at the origin, which is a stronger condition.
    – 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.
    – 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.
– 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.
– amd
Aug 15 at 19:48











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.






share|cite|improve this answer




















  • 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.
    – Sami Burton
    Aug 15 at 10:49










  • Would the use of the squeeze theorem be a better answer?
    – Sami Burton
    Aug 15 at 10:57










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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.






share|cite|improve this answer




















  • 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.
    – Sami Burton
    Aug 15 at 10:49










  • Would the use of the squeeze theorem be a better answer?
    – Sami Burton
    Aug 15 at 10:57














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.






share|cite|improve this answer




















  • 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.
    – Sami Burton
    Aug 15 at 10:49










  • Would the use of the squeeze theorem be a better answer?
    – Sami Burton
    Aug 15 at 10:57












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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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.
    – Sami Burton
    Aug 15 at 10:49










  • Would the use of the squeeze theorem be a better answer?
    – 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.
    – Sami Burton
    Aug 15 at 10:49










  • Would the use of the squeeze theorem be a better answer?
    – 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.
– 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.
– Sami Burton
Aug 15 at 10:49












Would the use of the squeeze theorem be a better answer?
– Sami Burton
Aug 15 at 10:57




Would the use of the squeeze theorem be a better answer?
– Sami Burton
Aug 15 at 10:57












 

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