The partial derivative
$begingroup$
Suppose the variables $x$ and $u$ are related by
$$x=u$$
Then I have a function $f=f(x)$ which does not explicitly depend on $u$.
Then is it true that $$frac{partial f}{partial u}=0$$?
partial-derivative
asked Dec 4 '18 at 2:19
glowstonetreesglowstonetrees
2,334418
$endgroup$
add a comment |
$begingroup$
Suppose the variables $x$ and $u$ are related by
$$x=u$$
Then I have a function $f=f(x)$ which does not explicitly depend on $u$.
Then is it true that $$frac{partial f}{partial u}=0$$?
partial-derivative
asked Dec 4 '18 at 2:19
glowstonetreesglowstonetrees
2,334418
$endgroup$
add a comment |
$begingroup$
Suppose the variables $x$ and $u$ are related by
$$x=u$$
Then I have a function $f=f(x)$ which does not explicitly depend on $u$.
Then is it true that $$frac{partial f}{partial u}=0$$?
partial-derivative
asked Dec 4 '18 at 2:19
glowstonetreesglowstonetrees
2,334418
$endgroup$
Suppose the variables $x$ and $u$ are related by
$$x=u$$
Then I have a function $f=f(x)$ which does not explicitly depend on $u$.
Then is it true that $$frac{partial f}{partial u}=0$$?
partial-derivative
partial-derivative
asked Dec 4 '18 at 2:19
glowstonetreesglowstonetrees
2,334418
asked Dec 4 '18 at 2:19
glowstonetreesglowstonetrees
2,334418
asked Dec 4 '18 at 2:19
glowstonetreesglowstonetrees
2,334418
asked Dec 4 '18 at 2:19
glowstonetreesglowstonetrees
2,334418
asked Dec 4 '18 at 2:19
glowstonetreesglowstonetrees
2,334418
2,334418
add a comment |
add a comment |
2 Answers
2
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$begingroup$
$frac{partial f}{partial u} = frac{partial f}{partial x} frac{partial x}{partial u} = f'(x)$, since $frac{partial x}{partial u} = 1$.
answered Dec 4 '18 at 2:24
Aditya DuaAditya Dua
1,00418
$endgroup$
add a comment |
$begingroup$
No, since you have only one variable, the symbol you should be using is
$$
frac{{rm d}}{{rm d}u}
$$
and
$$
require{cancel}
frac{{rm d}f}{{rm d}u} = frac{{rm d}f}{{rm d}x}cancelto{1}{frac{{rm d}x}{{rm d}u}} = frac{{rm d}f}{{rm d}x}
$$
answered Dec 4 '18 at 2:25
caveraccaverac
14.4k31130
$endgroup$
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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$begingroup$
$frac{partial f}{partial u} = frac{partial f}{partial x} frac{partial x}{partial u} = f'(x)$, since $frac{partial x}{partial u} = 1$.
answered Dec 4 '18 at 2:24
Aditya DuaAditya Dua
1,00418
$endgroup$
add a comment |
$begingroup$
$frac{partial f}{partial u} = frac{partial f}{partial x} frac{partial x}{partial u} = f'(x)$, since $frac{partial x}{partial u} = 1$.
answered Dec 4 '18 at 2:24
Aditya DuaAditya Dua
1,00418
$endgroup$
add a comment |
$begingroup$
$frac{partial f}{partial u} = frac{partial f}{partial x} frac{partial x}{partial u} = f'(x)$, since $frac{partial x}{partial u} = 1$.
answered Dec 4 '18 at 2:24
Aditya DuaAditya Dua
1,00418
$endgroup$
$frac{partial f}{partial u} = frac{partial f}{partial x} frac{partial x}{partial u} = f'(x)$, since $frac{partial x}{partial u} = 1$.
answered Dec 4 '18 at 2:24
Aditya DuaAditya Dua
1,00418
answered Dec 4 '18 at 2:24
Aditya DuaAditya Dua
1,00418
answered Dec 4 '18 at 2:24
Aditya DuaAditya Dua
1,00418
answered Dec 4 '18 at 2:24
Aditya DuaAditya Dua
1,00418
1,00418
add a comment |
add a comment |
$begingroup$
No, since you have only one variable, the symbol you should be using is
$$
frac{{rm d}}{{rm d}u}
$$
and
$$
require{cancel}
frac{{rm d}f}{{rm d}u} = frac{{rm d}f}{{rm d}x}cancelto{1}{frac{{rm d}x}{{rm d}u}} = frac{{rm d}f}{{rm d}x}
$$
answered Dec 4 '18 at 2:25
caveraccaverac
14.4k31130
$endgroup$
add a comment |
$begingroup$
No, since you have only one variable, the symbol you should be using is
$$
frac{{rm d}}{{rm d}u}
$$
and
$$
require{cancel}
frac{{rm d}f}{{rm d}u} = frac{{rm d}f}{{rm d}x}cancelto{1}{frac{{rm d}x}{{rm d}u}} = frac{{rm d}f}{{rm d}x}
$$
answered Dec 4 '18 at 2:25
caveraccaverac
14.4k31130
$endgroup$
add a comment |
$begingroup$
No, since you have only one variable, the symbol you should be using is
$$
frac{{rm d}}{{rm d}u}
$$
and
$$
require{cancel}
frac{{rm d}f}{{rm d}u} = frac{{rm d}f}{{rm d}x}cancelto{1}{frac{{rm d}x}{{rm d}u}} = frac{{rm d}f}{{rm d}x}
$$
answered Dec 4 '18 at 2:25
caveraccaverac
14.4k31130
$endgroup$
No, since you have only one variable, the symbol you should be using is
$$
frac{{rm d}}{{rm d}u}
$$
and
$$
require{cancel}
frac{{rm d}f}{{rm d}u} = frac{{rm d}f}{{rm d}x}cancelto{1}{frac{{rm d}x}{{rm d}u}} = frac{{rm d}f}{{rm d}x}
$$
answered Dec 4 '18 at 2:25
caveraccaverac
14.4k31130
answered Dec 4 '18 at 2:25
caveraccaverac
14.4k31130
answered Dec 4 '18 at 2:25
caveraccaverac
14.4k31130
answered Dec 4 '18 at 2:25
caveraccaverac
14.4k31130
14.4k31130
add a comment |
add a comment |
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