Second derivative at (0,0) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Finding directional derivatives at $(0,0)$ Multivariate calculusCheck whether the given function is differentiable at $(0,0)$Calculate the partial derivative at (0,0)Differentiability of piecewise functionsCheck whether second partial derivatives are equal at (0,0)Continuity and differentiability of $f(x,y)$ at $(0,0)$If $F(0,0)=0$ and $F(x,y)= fracxyx^2+y^2$ for $(x,y)neq (0,0)$ then $F$ is differentiable at $(0,0)$?What is the correct method to show that $f(x,y)$ is not-differentiable at $(0,0)$?Show that $g(x,y) = x^2y^2log(x^2+y^2), 0$ is differentiable in (0,0)Continuity of partial derivatives at (0,0)
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Second derivative at (0,0)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Finding directional derivatives at $(0,0)$ Multivariate calculusCheck whether the given function is differentiable at $(0,0)$Calculate the partial derivative at (0,0)Differentiability of piecewise functionsCheck whether second partial derivatives are equal at (0,0)Continuity and differentiability of $f(x,y)$ at $(0,0)$If $F(0,0)=0$ and $F(x,y)= fracxyx^2+y^2$ for $(x,y)neq (0,0)$ then $F$ is differentiable at $(0,0)$?What is the correct method to show that $f(x,y)$ is not-differentiable at $(0,0)$?Show that $g(x,y) = x^2y^2log(x^2+y^2), 0$ is differentiable in (0,0)Continuity of partial derivatives at (0,0)
$begingroup$
$$f(x,y)=left{beginmatrix
xyfracx^2-2y^2x^2+y^2 & (x,y)neq (0,0)\
0 & (x,y)=(0,0)
endmatrixright. $$ Calculate $ f_xy (0,0), f_yx(0,0) $.
- I found $f_x$ and then for $f_xy(0,0) $ I took the limit:
$$f_xy(0,0) =lim_yrightarrow 0fracf_x(0,y)-f_x(0,0)y-0$$
For $ f_x(0,0) $ I also took the limit:
$$f_x(0,0) =lim_xrightarrow 0fracf(x,0)-f(0,0)x-0$$
I got that $f_x(0,0) = 0 $ and $f_x(0,y) = -2y$ , so $$f_xy(0,0) = -2$$ - I did similar things for $f_yx(0,0) $ and I got that $$f_yx(0,0) = 0 $$
Can I be correct?
multivariable-calculus derivatives partial-derivative
asked Mar 25 at 19:40
Dr.MathematicsDr.Mathematics
516
$endgroup$
add a comment |
$begingroup$
$$f(x,y)=left{beginmatrix
xyfracx^2-2y^2x^2+y^2 & (x,y)neq (0,0)\
0 & (x,y)=(0,0)
endmatrixright. $$ Calculate $ f_xy (0,0), f_yx(0,0) $.
- I found $f_x$ and then for $f_xy(0,0) $ I took the limit:
$$f_xy(0,0) =lim_yrightarrow 0fracf_x(0,y)-f_x(0,0)y-0$$
For $ f_x(0,0) $ I also took the limit:
$$f_x(0,0) =lim_xrightarrow 0fracf(x,0)-f(0,0)x-0$$
I got that $f_x(0,0) = 0 $ and $f_x(0,y) = -2y$ , so $$f_xy(0,0) = -2$$ - I did similar things for $f_yx(0,0) $ and I got that $$f_yx(0,0) = 0 $$
Can I be correct?
multivariable-calculus derivatives partial-derivative
asked Mar 25 at 19:40
Dr.MathematicsDr.Mathematics
516
$endgroup$
$begingroup$
I get $f_yx(0,0)=1$ instead of $0$.
$endgroup$
– Ernie060
Mar 26 at 10:03
$begingroup$
@Ernie060 You are absolutely right! Thanks!
$endgroup$
– Dr.Mathematics
Mar 26 at 17:33
add a comment |
$begingroup$
$$f(x,y)=left{beginmatrix
xyfracx^2-2y^2x^2+y^2 & (x,y)neq (0,0)\
0 & (x,y)=(0,0)
endmatrixright. $$ Calculate $ f_xy (0,0), f_yx(0,0) $.
- I found $f_x$ and then for $f_xy(0,0) $ I took the limit:
$$f_xy(0,0) =lim_yrightarrow 0fracf_x(0,y)-f_x(0,0)y-0$$
For $ f_x(0,0) $ I also took the limit:
$$f_x(0,0) =lim_xrightarrow 0fracf(x,0)-f(0,0)x-0$$
I got that $f_x(0,0) = 0 $ and $f_x(0,y) = -2y$ , so $$f_xy(0,0) = -2$$ - I did similar things for $f_yx(0,0) $ and I got that $$f_yx(0,0) = 0 $$
Can I be correct?
multivariable-calculus derivatives partial-derivative
asked Mar 25 at 19:40
Dr.MathematicsDr.Mathematics
516
$endgroup$
$$f(x,y)=left{beginmatrix
xyfracx^2-2y^2x^2+y^2 & (x,y)neq (0,0)\
0 & (x,y)=(0,0)
endmatrixright. $$ Calculate $ f_xy (0,0), f_yx(0,0) $.
- I found $f_x$ and then for $f_xy(0,0) $ I took the limit:
$$f_xy(0,0) =lim_yrightarrow 0fracf_x(0,y)-f_x(0,0)y-0$$
For $ f_x(0,0) $ I also took the limit:
$$f_x(0,0) =lim_xrightarrow 0fracf(x,0)-f(0,0)x-0$$
I got that $f_x(0,0) = 0 $ and $f_x(0,y) = -2y$ , so $$f_xy(0,0) = -2$$ - I did similar things for $f_yx(0,0) $ and I got that $$f_yx(0,0) = 0 $$
Can I be correct?
multivariable-calculus derivatives partial-derivative
multivariable-calculus derivatives partial-derivative
asked Mar 25 at 19:40
Dr.MathematicsDr.Mathematics
516
asked Mar 25 at 19:40
Dr.MathematicsDr.Mathematics
516
asked Mar 25 at 19:40
Dr.MathematicsDr.Mathematics
516
asked Mar 25 at 19:40
Dr.MathematicsDr.Mathematics
516
asked Mar 25 at 19:40
Dr.MathematicsDr.Mathematics
516
516
$begingroup$
I get $f_yx(0,0)=1$ instead of $0$.
$endgroup$
– Ernie060
Mar 26 at 10:03
$begingroup$
@Ernie060 You are absolutely right! Thanks!
$endgroup$
– Dr.Mathematics
Mar 26 at 17:33
add a comment |
$begingroup$
I get $f_yx(0,0)=1$ instead of $0$.
$endgroup$
– Ernie060
Mar 26 at 10:03
$begingroup$
@Ernie060 You are absolutely right! Thanks!
$endgroup$
– Dr.Mathematics
Mar 26 at 17:33
$begingroup$
I get $f_yx(0,0)=1$ instead of $0$.
$endgroup$
– Ernie060
Mar 26 at 10:03
$begingroup$
I get $f_yx(0,0)=1$ instead of $0$.
$endgroup$
– Ernie060
Mar 26 at 10:03
$begingroup$
@Ernie060 You are absolutely right! Thanks!
$endgroup$
– Dr.Mathematics
Mar 26 at 17:33
$begingroup$
@Ernie060 You are absolutely right! Thanks!
$endgroup$
– Dr.Mathematics
Mar 26 at 17:33
add a comment |
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$begingroup$
I get $f_yx(0,0)=1$ instead of $0$.
$endgroup$
– Ernie060
Mar 26 at 10:03
$begingroup$
@Ernie060 You are absolutely right! Thanks!
$endgroup$
– Dr.Mathematics
Mar 26 at 17:33