differentiability in the origin of f(x,y)
I have $$
f(x,y) = cases{ sqrt{xy}& if $x>0,y>0$ \
-sqrt{xy}& if $x<0,y<0$ \
0 }
$$
I want calculate directional derivative $D_vf(0,0)$ with $v=(1,1)$
f is not differentiable in the origin but do directional derivatives exist ?
real-analysis multivariable-calculus
edited Dec 4 at 11:22
José Carlos Santos
149k22117219
asked Nov 26 at 18:22
Giulia B.
415211
add a comment |
I have $$
f(x,y) = cases{ sqrt{xy}& if $x>0,y>0$ \
-sqrt{xy}& if $x<0,y<0$ \
0 }
$$
I want calculate directional derivative $D_vf(0,0)$ with $v=(1,1)$
f is not differentiable in the origin but do directional derivatives exist ?
real-analysis multivariable-calculus
edited Dec 4 at 11:22
José Carlos Santos
149k22117219
asked Nov 26 at 18:22
Giulia B.
415211
add a comment |
I have $$
f(x,y) = cases{ sqrt{xy}& if $x>0,y>0$ \
-sqrt{xy}& if $x<0,y<0$ \
0 }
$$
I want calculate directional derivative $D_vf(0,0)$ with $v=(1,1)$
f is not differentiable in the origin but do directional derivatives exist ?
real-analysis multivariable-calculus
edited Dec 4 at 11:22
José Carlos Santos
149k22117219
asked Nov 26 at 18:22
Giulia B.
415211
I have $$
f(x,y) = cases{ sqrt{xy}& if $x>0,y>0$ \
-sqrt{xy}& if $x<0,y<0$ \
0 }
$$
I want calculate directional derivative $D_vf(0,0)$ with $v=(1,1)$
f is not differentiable in the origin but do directional derivatives exist ?
real-analysis multivariable-calculus
real-analysis multivariable-calculus
edited Dec 4 at 11:22
José Carlos Santos
149k22117219
asked Nov 26 at 18:22
Giulia B.
415211
edited Dec 4 at 11:22
José Carlos Santos
149k22117219
asked Nov 26 at 18:22
Giulia B.
415211
edited Dec 4 at 11:22
José Carlos Santos
149k22117219
edited Dec 4 at 11:22
José Carlos Santos
149k22117219
edited Dec 4 at 11:22
José Carlos Santos
149k22117219
149k22117219
asked Nov 26 at 18:22
Giulia B.
415211
asked Nov 26 at 18:22
Giulia B.
415211
asked Nov 26 at 18:22
Giulia B.
415211
415211
add a comment |
add a comment |
1 Answer
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Hint: Your question is: does the limit $displaystylelim_{tto0}frac{f(t,t)}t$ exist? What do you think?
answered Nov 26 at 18:25
José Carlos Santos
149k22117219
$lim_{tto0^+}frac{f(t,t)}t=lim_{tto0^+}frac{|t|}t=1$ and $lim_{tto0^-}frac{f(t,t)}t=lim_{tto0^-}frac{-|t|}t=1$ so exist?
– Giulia B.
Nov 26 at 18:33
Yes, it does, What's the doubt?
– José Carlos Santos
Nov 26 at 18:36
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint: Your question is: does the limit $displaystylelim_{tto0}frac{f(t,t)}t$ exist? What do you think?
answered Nov 26 at 18:25
José Carlos Santos
149k22117219
$lim_{tto0^+}frac{f(t,t)}t=lim_{tto0^+}frac{|t|}t=1$ and $lim_{tto0^-}frac{f(t,t)}t=lim_{tto0^-}frac{-|t|}t=1$ so exist?
– Giulia B.
Nov 26 at 18:33
Yes, it does, What's the doubt?
– José Carlos Santos
Nov 26 at 18:36
add a comment |
Hint: Your question is: does the limit $displaystylelim_{tto0}frac{f(t,t)}t$ exist? What do you think?
answered Nov 26 at 18:25
José Carlos Santos
149k22117219
$lim_{tto0^+}frac{f(t,t)}t=lim_{tto0^+}frac{|t|}t=1$ and $lim_{tto0^-}frac{f(t,t)}t=lim_{tto0^-}frac{-|t|}t=1$ so exist?
– Giulia B.
Nov 26 at 18:33
Yes, it does, What's the doubt?
– José Carlos Santos
Nov 26 at 18:36
add a comment |
Hint: Your question is: does the limit $displaystylelim_{tto0}frac{f(t,t)}t$ exist? What do you think?
answered Nov 26 at 18:25
José Carlos Santos
149k22117219
Hint: Your question is: does the limit $displaystylelim_{tto0}frac{f(t,t)}t$ exist? What do you think?
answered Nov 26 at 18:25
José Carlos Santos
149k22117219
answered Nov 26 at 18:25
José Carlos Santos
149k22117219
answered Nov 26 at 18:25
José Carlos Santos
149k22117219
answered Nov 26 at 18:25
José Carlos Santos
149k22117219
149k22117219
$lim_{tto0^+}frac{f(t,t)}t=lim_{tto0^+}frac{|t|}t=1$ and $lim_{tto0^-}frac{f(t,t)}t=lim_{tto0^-}frac{-|t|}t=1$ so exist?
– Giulia B.
Nov 26 at 18:33
Yes, it does, What's the doubt?
– José Carlos Santos
Nov 26 at 18:36
add a comment |
$lim_{tto0^+}frac{f(t,t)}t=lim_{tto0^+}frac{|t|}t=1$ and $lim_{tto0^-}frac{f(t,t)}t=lim_{tto0^-}frac{-|t|}t=1$ so exist?
– Giulia B.
Nov 26 at 18:33
Yes, it does, What's the doubt?
– José Carlos Santos
Nov 26 at 18:36
$lim_{tto0^+}frac{f(t,t)}t=lim_{tto0^+}frac{|t|}t=1$ and $lim_{tto0^-}frac{f(t,t)}t=lim_{tto0^-}frac{-|t|}t=1$ so exist?
– Giulia B.
Nov 26 at 18:33
$lim_{tto0^+}frac{f(t,t)}t=lim_{tto0^+}frac{|t|}t=1$ and $lim_{tto0^-}frac{f(t,t)}t=lim_{tto0^-}frac{-|t|}t=1$ so exist?
– Giulia B.
Nov 26 at 18:33
Yes, it does, What's the doubt?
– José Carlos Santos
Nov 26 at 18:36
Yes, it does, What's the doubt?
– José Carlos Santos
Nov 26 at 18:36
add a comment |
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