Directional derivative and manipulation of nonlinear functions
$begingroup$
I have a question about a few manipulations when computing a directional derivative at the point $(0,0)$.
We have $f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h}$ where $f(x,y) = (xy)^frac{1}{3}$ and $u = (u_1, u_2)$. This is just the definition of a directional derivative.
Then we expand this to: $f'(0,u) = limlimits_{h to 0} frac{(h u_1h u_2)^frac{1}{3}}{h}$
I don't understand the steps of this fully. Are the precise steps:
$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}$
This seems to make sense to me but I just want to make sure I am allowed to do these operations inside the function (in particular, taking the scalar $h$ inside in the last step).
Thanks for your help.
calculus derivatives
asked Dec 6 '18 at 0:06
ChristianChristian
388
$endgroup$
add a comment |
$begingroup$
I have a question about a few manipulations when computing a directional derivative at the point $(0,0)$.
We have $f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h}$ where $f(x,y) = (xy)^frac{1}{3}$ and $u = (u_1, u_2)$. This is just the definition of a directional derivative.
Then we expand this to: $f'(0,u) = limlimits_{h to 0} frac{(h u_1h u_2)^frac{1}{3}}{h}$
I don't understand the steps of this fully. Are the precise steps:
$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}$
This seems to make sense to me but I just want to make sure I am allowed to do these operations inside the function (in particular, taking the scalar $h$ inside in the last step).
Thanks for your help.
calculus derivatives
asked Dec 6 '18 at 0:06
ChristianChristian
388
$endgroup$
add a comment |
$begingroup$
I have a question about a few manipulations when computing a directional derivative at the point $(0,0)$.
We have $f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h}$ where $f(x,y) = (xy)^frac{1}{3}$ and $u = (u_1, u_2)$. This is just the definition of a directional derivative.
Then we expand this to: $f'(0,u) = limlimits_{h to 0} frac{(h u_1h u_2)^frac{1}{3}}{h}$
I don't understand the steps of this fully. Are the precise steps:
$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}$
This seems to make sense to me but I just want to make sure I am allowed to do these operations inside the function (in particular, taking the scalar $h$ inside in the last step).
Thanks for your help.
calculus derivatives
asked Dec 6 '18 at 0:06
ChristianChristian
388
$endgroup$
I have a question about a few manipulations when computing a directional derivative at the point $(0,0)$.
We have $f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h}$ where $f(x,y) = (xy)^frac{1}{3}$ and $u = (u_1, u_2)$. This is just the definition of a directional derivative.
Then we expand this to: $f'(0,u) = limlimits_{h to 0} frac{(h u_1h u_2)^frac{1}{3}}{h}$
I don't understand the steps of this fully. Are the precise steps:
$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}$
This seems to make sense to me but I just want to make sure I am allowed to do these operations inside the function (in particular, taking the scalar $h$ inside in the last step).
Thanks for your help.
calculus derivatives
calculus derivatives
asked Dec 6 '18 at 0:06
ChristianChristian
388
asked Dec 6 '18 at 0:06
ChristianChristian
388
asked Dec 6 '18 at 0:06
ChristianChristian
388
asked Dec 6 '18 at 0:06
ChristianChristian
388
asked Dec 6 '18 at 0:06
ChristianChristian
388
388
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes we have
$$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}=limlimits_{h to 0} frac{h^frac23(u_1u_2)^frac13}{h}$$
and for $u_1,u_2 neq 0$ the directional derivative doesn't exist.
answered Dec 6 '18 at 0:10
gimusigimusi
92.9k84494
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Yes we have
$$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}=limlimits_{h to 0} frac{h^frac23(u_1u_2)^frac13}{h}$$
and for $u_1,u_2 neq 0$ the directional derivative doesn't exist.
answered Dec 6 '18 at 0:10
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
Yes we have
$$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}=limlimits_{h to 0} frac{h^frac23(u_1u_2)^frac13}{h}$$
and for $u_1,u_2 neq 0$ the directional derivative doesn't exist.
answered Dec 6 '18 at 0:10
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
Yes we have
$$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}=limlimits_{h to 0} frac{h^frac23(u_1u_2)^frac13}{h}$$
and for $u_1,u_2 neq 0$ the directional derivative doesn't exist.
answered Dec 6 '18 at 0:10
gimusigimusi
92.9k84494
$endgroup$
Yes we have
$$f'(0,u) = limlimits_{h to 0} frac{f(hu)}{h} = limlimits_{h to 0} frac{f(h(u_1,u_2))}{h}=limlimits_{h to 0} frac{f((hu_1,hu_2))}{h}=limlimits_{h to 0} frac{h^frac23(u_1u_2)^frac13}{h}$$
and for $u_1,u_2 neq 0$ the directional derivative doesn't exist.
answered Dec 6 '18 at 0:10
gimusigimusi
92.9k84494
answered Dec 6 '18 at 0:10
gimusigimusi
92.9k84494
answered Dec 6 '18 at 0:10
gimusigimusi
92.9k84494
answered Dec 6 '18 at 0:10
gimusigimusi
92.9k84494
92.9k84494
add a comment |
add a comment |
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