Mixing
Why Does This Partial Derivative Exist Everywhere?
$begingroup$
The following function is taken from my textbook example.
begin{cases}
f(x,y)=frac{2xy^2}{x^2+y^4}, &(x,y) neq (0,0)\
f(x,y)=0, &(x,y)=(0,0)
end{cases}
My textbook asserts that the partial derivatives of $f$ exists everywhere, which I do not understand. I tried to solve this myself, but it seems like I am misunderstanding something because my work shows some inconcsistencies:
To calculate $f_x$ at $(0,0)$:
$$
lim_{x to 0}frac{f(x,y)-f(0,0)}{x}=lim_{x to 0} frac{2y^2}{x^2+y^4}=frac{2}{y^2} to infty text{ as } y to 0
$$
However, if I fix $y=0 to f(x,0)=0 space forall x$, then:
$$
lim_{x to 0, y=0}frac{f(x,0)-f(0,0)}{x}=0
$$
Does this mean $f_x$ does not exist at $(0,0)$? But this contradicts with my textbook's claim.
limits continuity partial-derivative
asked Jan 19 at 8:45
A Slow LearnerA Slow Learner
453212
$endgroup$
add a comment |
$begingroup$
The following function is taken from my textbook example.
begin{cases}
f(x,y)=frac{2xy^2}{x^2+y^4}, &(x,y) neq (0,0)\
f(x,y)=0, &(x,y)=(0,0)
end{cases}
My textbook asserts that the partial derivatives of $f$ exists everywhere, which I do not understand. I tried to solve this myself, but it seems like I am misunderstanding something because my work shows some inconcsistencies:
To calculate $f_x$ at $(0,0)$:
$$
lim_{x to 0}frac{f(x,y)-f(0,0)}{x}=lim_{x to 0} frac{2y^2}{x^2+y^4}=frac{2}{y^2} to infty text{ as } y to 0
$$
However, if I fix $y=0 to f(x,0)=0 space forall x$, then:
$$
lim_{x to 0, y=0}frac{f(x,0)-f(0,0)}{x}=0
$$
Does this mean $f_x$ does not exist at $(0,0)$? But this contradicts with my textbook's claim.
limits continuity partial-derivative
asked Jan 19 at 8:45
A Slow LearnerA Slow Learner
453212
$endgroup$
add a comment |
$begingroup$
The following function is taken from my textbook example.
begin{cases}
f(x,y)=frac{2xy^2}{x^2+y^4}, &(x,y) neq (0,0)\
f(x,y)=0, &(x,y)=(0,0)
end{cases}
My textbook asserts that the partial derivatives of $f$ exists everywhere, which I do not understand. I tried to solve this myself, but it seems like I am misunderstanding something because my work shows some inconcsistencies:
To calculate $f_x$ at $(0,0)$:
$$
lim_{x to 0}frac{f(x,y)-f(0,0)}{x}=lim_{x to 0} frac{2y^2}{x^2+y^4}=frac{2}{y^2} to infty text{ as } y to 0
$$
However, if I fix $y=0 to f(x,0)=0 space forall x$, then:
$$
lim_{x to 0, y=0}frac{f(x,0)-f(0,0)}{x}=0
$$
Does this mean $f_x$ does not exist at $(0,0)$? But this contradicts with my textbook's claim.
limits continuity partial-derivative
asked Jan 19 at 8:45
A Slow LearnerA Slow Learner
453212
$endgroup$
The following function is taken from my textbook example.
begin{cases}
f(x,y)=frac{2xy^2}{x^2+y^4}, &(x,y) neq (0,0)\
f(x,y)=0, &(x,y)=(0,0)
end{cases}
My textbook asserts that the partial derivatives of $f$ exists everywhere, which I do not understand. I tried to solve this myself, but it seems like I am misunderstanding something because my work shows some inconcsistencies:
To calculate $f_x$ at $(0,0)$:
$$
lim_{x to 0}frac{f(x,y)-f(0,0)}{x}=lim_{x to 0} frac{2y^2}{x^2+y^4}=frac{2}{y^2} to infty text{ as } y to 0
$$
However, if I fix $y=0 to f(x,0)=0 space forall x$, then:
$$
lim_{x to 0, y=0}frac{f(x,0)-f(0,0)}{x}=0
$$
Does this mean $f_x$ does not exist at $(0,0)$? But this contradicts with my textbook's claim.
limits continuity partial-derivative
limits continuity partial-derivative
asked Jan 19 at 8:45
A Slow LearnerA Slow Learner
453212
asked Jan 19 at 8:45
A Slow LearnerA Slow Learner
453212
asked Jan 19 at 8:45
A Slow LearnerA Slow Learner
453212
asked Jan 19 at 8:45
A Slow LearnerA Slow Learner
453212
asked Jan 19 at 8:45
A Slow LearnerA Slow Learner
453212
453212
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For $f_x(0)$ you must compute $lim_{x to 0} frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis); the first computation is irrelevant. It's clear that $f(x,0) = 0$ for all $x neq 0$ from the formula and also $f(0,0)=0$, hence the limit and the partial derivative is just $0$.
Similarly the other partial derivative is $lim_{y to 0} frac{f(0,y) - f(0,0)}{y}$, which is similarly seen to be $0$ too.
answered Jan 19 at 9:12
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
"For $f_x(0)$ you must compute $lim_{x to 0} frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis)". Does this mean I can also fix, for instance, $y=3$, to compute $f_x$? This is essentially what the first calculation did: I wanted to approach (0,0) from an arbitrary horizontal line.
$endgroup$
– A Slow Learner
Jan 19 at 9:53
$begingroup$
@ASlowLearner If you take $y=3$ and the limit as $x$ tends to $0$ we have the partial derivative at $(0,3)$ in the $x$-direction. The partial derivative at points not $(0,0)$ can be computed by the formula. Only at $(0,0)$ do we need the explicit limits.
$endgroup$
– Henno Brandsma
Jan 19 at 11:00
$begingroup$
so the only way to compute $f_x(0,0)$ is to set $y=0$?
$endgroup$
– A Slow Learner
Jan 19 at 11:20
$begingroup$
@ASlowLearner indeed.
$endgroup$
– Henno Brandsma
Jan 19 at 11:21
$begingroup$
Ok. But looking back in my post, where $frac{2}{y^2} to 0$ as $y to 0$. This limit should be equal to $0$ is $f_x(0,0) = 0$, as done in the second calculation. But it is not. Am I still missing something?
$endgroup$
– A Slow Learner
Jan 19 at 11:25
|
show 2 more comments
$begingroup$
To compute the partial derivative of $f$ in $(x_0, y_0)$ with respect to $x$, you only vary $x$ around $x_0$ and keep $y=y_0$ the whole time.
Your second calculation is in fact $f_x(0,0)$: you fix $y=0$ and see how the differences behave when you send $x$ to $0$. So you've shown that $f_x$ exists in $(0,0)$.
The first calculation isn't anything standard, because you're varying $x$ but your $y$ is not fixed: you have $y=y$ in one term and $y=0$ in the other. For instance, changing $lim_{x to 0} frac{f(x,y) - f(0,0)}{x}$ to $lim_{x to 0} frac{f(x,y) - f(0,y)}{x}$ would give you $f_x(0,y)$ for arbitrary $y ne 0$.
Taking the limit of that for $y to 0$ gives you $lim_{y to 0} f_x(0,y)$.
Note: this change does not affect the result, though. So you still end up getting $lim_{y to 0} f_x(0,y) = infty$. That gives you insight into the (dis-)continuity of $f_x$.
answered Jan 19 at 9:19
yawumayawuma
11
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add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
For $f_x(0)$ you must compute $lim_{x to 0} frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis); the first computation is irrelevant. It's clear that $f(x,0) = 0$ for all $x neq 0$ from the formula and also $f(0,0)=0$, hence the limit and the partial derivative is just $0$.
Similarly the other partial derivative is $lim_{y to 0} frac{f(0,y) - f(0,0)}{y}$, which is similarly seen to be $0$ too.
answered Jan 19 at 9:12
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
"For $f_x(0)$ you must compute $lim_{x to 0} frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis)". Does this mean I can also fix, for instance, $y=3$, to compute $f_x$? This is essentially what the first calculation did: I wanted to approach (0,0) from an arbitrary horizontal line.
$endgroup$
– A Slow Learner
Jan 19 at 9:53
$begingroup$
@ASlowLearner If you take $y=3$ and the limit as $x$ tends to $0$ we have the partial derivative at $(0,3)$ in the $x$-direction. The partial derivative at points not $(0,0)$ can be computed by the formula. Only at $(0,0)$ do we need the explicit limits.
$endgroup$
– Henno Brandsma
Jan 19 at 11:00
$begingroup$
so the only way to compute $f_x(0,0)$ is to set $y=0$?
$endgroup$
– A Slow Learner
Jan 19 at 11:20
$begingroup$
@ASlowLearner indeed.
$endgroup$
– Henno Brandsma
Jan 19 at 11:21
$begingroup$
Ok. But looking back in my post, where $frac{2}{y^2} to 0$ as $y to 0$. This limit should be equal to $0$ is $f_x(0,0) = 0$, as done in the second calculation. But it is not. Am I still missing something?
$endgroup$
– A Slow Learner
Jan 19 at 11:25
|
show 2 more comments
$begingroup$
For $f_x(0)$ you must compute $lim_{x to 0} frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis); the first computation is irrelevant. It's clear that $f(x,0) = 0$ for all $x neq 0$ from the formula and also $f(0,0)=0$, hence the limit and the partial derivative is just $0$.
Similarly the other partial derivative is $lim_{y to 0} frac{f(0,y) - f(0,0)}{y}$, which is similarly seen to be $0$ too.
answered Jan 19 at 9:12
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
"For $f_x(0)$ you must compute $lim_{x to 0} frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis)". Does this mean I can also fix, for instance, $y=3$, to compute $f_x$? This is essentially what the first calculation did: I wanted to approach (0,0) from an arbitrary horizontal line.
$endgroup$
– A Slow Learner
Jan 19 at 9:53
$begingroup$
@ASlowLearner If you take $y=3$ and the limit as $x$ tends to $0$ we have the partial derivative at $(0,3)$ in the $x$-direction. The partial derivative at points not $(0,0)$ can be computed by the formula. Only at $(0,0)$ do we need the explicit limits.
$endgroup$
– Henno Brandsma
Jan 19 at 11:00
$begingroup$
so the only way to compute $f_x(0,0)$ is to set $y=0$?
$endgroup$
– A Slow Learner
Jan 19 at 11:20
$begingroup$
@ASlowLearner indeed.
$endgroup$
– Henno Brandsma
Jan 19 at 11:21
$begingroup$
Ok. But looking back in my post, where $frac{2}{y^2} to 0$ as $y to 0$. This limit should be equal to $0$ is $f_x(0,0) = 0$, as done in the second calculation. But it is not. Am I still missing something?
$endgroup$
– A Slow Learner
Jan 19 at 11:25
|
show 2 more comments
$begingroup$
For $f_x(0)$ you must compute $lim_{x to 0} frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis); the first computation is irrelevant. It's clear that $f(x,0) = 0$ for all $x neq 0$ from the formula and also $f(0,0)=0$, hence the limit and the partial derivative is just $0$.
Similarly the other partial derivative is $lim_{y to 0} frac{f(0,y) - f(0,0)}{y}$, which is similarly seen to be $0$ too.
answered Jan 19 at 9:12
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
For $f_x(0)$ you must compute $lim_{x to 0} frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis); the first computation is irrelevant. It's clear that $f(x,0) = 0$ for all $x neq 0$ from the formula and also $f(0,0)=0$, hence the limit and the partial derivative is just $0$.
Similarly the other partial derivative is $lim_{y to 0} frac{f(0,y) - f(0,0)}{y}$, which is similarly seen to be $0$ too.
answered Jan 19 at 9:12
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 19 at 9:12
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 19 at 9:12
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 19 at 9:12
Henno BrandsmaHenno Brandsma
111k348118
111k348118
$begingroup$
"For $f_x(0)$ you must compute $lim_{x to 0} frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis)". Does this mean I can also fix, for instance, $y=3$, to compute $f_x$? This is essentially what the first calculation did: I wanted to approach (0,0) from an arbitrary horizontal line.
$endgroup$
– A Slow Learner
Jan 19 at 9:53
$begingroup$
@ASlowLearner If you take $y=3$ and the limit as $x$ tends to $0$ we have the partial derivative at $(0,3)$ in the $x$-direction. The partial derivative at points not $(0,0)$ can be computed by the formula. Only at $(0,0)$ do we need the explicit limits.
$endgroup$
– Henno Brandsma
Jan 19 at 11:00
$begingroup$
so the only way to compute $f_x(0,0)$ is to set $y=0$?
$endgroup$
– A Slow Learner
Jan 19 at 11:20
$begingroup$
@ASlowLearner indeed.
$endgroup$
– Henno Brandsma
Jan 19 at 11:21
$begingroup$
Ok. But looking back in my post, where $frac{2}{y^2} to 0$ as $y to 0$. This limit should be equal to $0$ is $f_x(0,0) = 0$, as done in the second calculation. But it is not. Am I still missing something?
$endgroup$
– A Slow Learner
Jan 19 at 11:25
|
show 2 more comments
$begingroup$
"For $f_x(0)$ you must compute $lim_{x to 0} frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis)". Does this mean I can also fix, for instance, $y=3$, to compute $f_x$? This is essentially what the first calculation did: I wanted to approach (0,0) from an arbitrary horizontal line.
$endgroup$
– A Slow Learner
Jan 19 at 9:53
$begingroup$
@ASlowLearner If you take $y=3$ and the limit as $x$ tends to $0$ we have the partial derivative at $(0,3)$ in the $x$-direction. The partial derivative at points not $(0,0)$ can be computed by the formula. Only at $(0,0)$ do we need the explicit limits.
$endgroup$
– Henno Brandsma
Jan 19 at 11:00
$begingroup$
so the only way to compute $f_x(0,0)$ is to set $y=0$?
$endgroup$
– A Slow Learner
Jan 19 at 11:20
$begingroup$
@ASlowLearner indeed.
$endgroup$
– Henno Brandsma
Jan 19 at 11:21
$begingroup$
Ok. But looking back in my post, where $frac{2}{y^2} to 0$ as $y to 0$. This limit should be equal to $0$ is $f_x(0,0) = 0$, as done in the second calculation. But it is not. Am I still missing something?
$endgroup$
– A Slow Learner
Jan 19 at 11:25
$begingroup$
"For $f_x(0)$ you must compute $lim_{x to 0} frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis)". Does this mean I can also fix, for instance, $y=3$, to compute $f_x$? This is essentially what the first calculation did: I wanted to approach (0,0) from an arbitrary horizontal line.
$endgroup$
– A Slow Learner
Jan 19 at 9:53
$begingroup$
"For $f_x(0)$ you must compute $lim_{x to 0} frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis)". Does this mean I can also fix, for instance, $y=3$, to compute $f_x$? This is essentially what the first calculation did: I wanted to approach (0,0) from an arbitrary horizontal line.
$endgroup$
– A Slow Learner
Jan 19 at 9:53
$begingroup$
@ASlowLearner If you take $y=3$ and the limit as $x$ tends to $0$ we have the partial derivative at $(0,3)$ in the $x$-direction. The partial derivative at points not $(0,0)$ can be computed by the formula. Only at $(0,0)$ do we need the explicit limits.
$endgroup$
– Henno Brandsma
Jan 19 at 11:00
$begingroup$
@ASlowLearner If you take $y=3$ and the limit as $x$ tends to $0$ we have the partial derivative at $(0,3)$ in the $x$-direction. The partial derivative at points not $(0,0)$ can be computed by the formula. Only at $(0,0)$ do we need the explicit limits.
$endgroup$
– Henno Brandsma
Jan 19 at 11:00
$begingroup$
so the only way to compute $f_x(0,0)$ is to set $y=0$?
$endgroup$
– A Slow Learner
Jan 19 at 11:20
$begingroup$
so the only way to compute $f_x(0,0)$ is to set $y=0$?
$endgroup$
– A Slow Learner
Jan 19 at 11:20
$begingroup$
@ASlowLearner indeed.
$endgroup$
– Henno Brandsma
Jan 19 at 11:21
$begingroup$
@ASlowLearner indeed.
$endgroup$
– Henno Brandsma
Jan 19 at 11:21
$begingroup$
Ok. But looking back in my post, where $frac{2}{y^2} to 0$ as $y to 0$. This limit should be equal to $0$ is $f_x(0,0) = 0$, as done in the second calculation. But it is not. Am I still missing something?
$endgroup$
– A Slow Learner
Jan 19 at 11:25
$begingroup$
Ok. But looking back in my post, where $frac{2}{y^2} to 0$ as $y to 0$. This limit should be equal to $0$ is $f_x(0,0) = 0$, as done in the second calculation. But it is not. Am I still missing something?
$endgroup$
– A Slow Learner
Jan 19 at 11:25
|
show 2 more comments
$begingroup$
To compute the partial derivative of $f$ in $(x_0, y_0)$ with respect to $x$, you only vary $x$ around $x_0$ and keep $y=y_0$ the whole time.
Your second calculation is in fact $f_x(0,0)$: you fix $y=0$ and see how the differences behave when you send $x$ to $0$. So you've shown that $f_x$ exists in $(0,0)$.
The first calculation isn't anything standard, because you're varying $x$ but your $y$ is not fixed: you have $y=y$ in one term and $y=0$ in the other. For instance, changing $lim_{x to 0} frac{f(x,y) - f(0,0)}{x}$ to $lim_{x to 0} frac{f(x,y) - f(0,y)}{x}$ would give you $f_x(0,y)$ for arbitrary $y ne 0$.
Taking the limit of that for $y to 0$ gives you $lim_{y to 0} f_x(0,y)$.
Note: this change does not affect the result, though. So you still end up getting $lim_{y to 0} f_x(0,y) = infty$. That gives you insight into the (dis-)continuity of $f_x$.
answered Jan 19 at 9:19
yawumayawuma
11
$endgroup$
add a comment |
$begingroup$
To compute the partial derivative of $f$ in $(x_0, y_0)$ with respect to $x$, you only vary $x$ around $x_0$ and keep $y=y_0$ the whole time.
Your second calculation is in fact $f_x(0,0)$: you fix $y=0$ and see how the differences behave when you send $x$ to $0$. So you've shown that $f_x$ exists in $(0,0)$.
The first calculation isn't anything standard, because you're varying $x$ but your $y$ is not fixed: you have $y=y$ in one term and $y=0$ in the other. For instance, changing $lim_{x to 0} frac{f(x,y) - f(0,0)}{x}$ to $lim_{x to 0} frac{f(x,y) - f(0,y)}{x}$ would give you $f_x(0,y)$ for arbitrary $y ne 0$.
Taking the limit of that for $y to 0$ gives you $lim_{y to 0} f_x(0,y)$.
Note: this change does not affect the result, though. So you still end up getting $lim_{y to 0} f_x(0,y) = infty$. That gives you insight into the (dis-)continuity of $f_x$.
answered Jan 19 at 9:19
yawumayawuma
11
$endgroup$
add a comment |
$begingroup$
To compute the partial derivative of $f$ in $(x_0, y_0)$ with respect to $x$, you only vary $x$ around $x_0$ and keep $y=y_0$ the whole time.
Your second calculation is in fact $f_x(0,0)$: you fix $y=0$ and see how the differences behave when you send $x$ to $0$. So you've shown that $f_x$ exists in $(0,0)$.
The first calculation isn't anything standard, because you're varying $x$ but your $y$ is not fixed: you have $y=y$ in one term and $y=0$ in the other. For instance, changing $lim_{x to 0} frac{f(x,y) - f(0,0)}{x}$ to $lim_{x to 0} frac{f(x,y) - f(0,y)}{x}$ would give you $f_x(0,y)$ for arbitrary $y ne 0$.
Taking the limit of that for $y to 0$ gives you $lim_{y to 0} f_x(0,y)$.
Note: this change does not affect the result, though. So you still end up getting $lim_{y to 0} f_x(0,y) = infty$. That gives you insight into the (dis-)continuity of $f_x$.
answered Jan 19 at 9:19
yawumayawuma
11
$endgroup$
To compute the partial derivative of $f$ in $(x_0, y_0)$ with respect to $x$, you only vary $x$ around $x_0$ and keep $y=y_0$ the whole time.
Your second calculation is in fact $f_x(0,0)$: you fix $y=0$ and see how the differences behave when you send $x$ to $0$. So you've shown that $f_x$ exists in $(0,0)$.
The first calculation isn't anything standard, because you're varying $x$ but your $y$ is not fixed: you have $y=y$ in one term and $y=0$ in the other. For instance, changing $lim_{x to 0} frac{f(x,y) - f(0,0)}{x}$ to $lim_{x to 0} frac{f(x,y) - f(0,y)}{x}$ would give you $f_x(0,y)$ for arbitrary $y ne 0$.
Taking the limit of that for $y to 0$ gives you $lim_{y to 0} f_x(0,y)$.
Note: this change does not affect the result, though. So you still end up getting $lim_{y to 0} f_x(0,y) = infty$. That gives you insight into the (dis-)continuity of $f_x$.
answered Jan 19 at 9:19
yawumayawuma
11
answered Jan 19 at 9:19
yawumayawuma
11
answered Jan 19 at 9:19
yawumayawuma
11
answered Jan 19 at 9:19
yawumayawuma
11
11
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