If $f = f(x,y)$ and $C$ is constant, then is this true: $frac{partial f}{partial (Cx)}=frac{1}{C}frac{partial...
$begingroup$
Suppose $f=f(x,y)$ is any function of 2 variables and $C=mbox{const}$. Can one do
$$frac{partial f}{partial (Cx)}=frac{1}{C}frac{partial f}{partial x}$$
Is this always correct? Or do we need the derivative $f_x'$ to exist, or be continuous before we can write the above?
multivariable-calculus derivatives partial-derivative
asked Dec 18 '18 at 15:56
user142523user142523
18112
$endgroup$
add a comment |
$begingroup$
Suppose $f=f(x,y)$ is any function of 2 variables and $C=mbox{const}$. Can one do
$$frac{partial f}{partial (Cx)}=frac{1}{C}frac{partial f}{partial x}$$
Is this always correct? Or do we need the derivative $f_x'$ to exist, or be continuous before we can write the above?
multivariable-calculus derivatives partial-derivative
asked Dec 18 '18 at 15:56
user142523user142523
18112
$endgroup$
add a comment |
$begingroup$
Suppose $f=f(x,y)$ is any function of 2 variables and $C=mbox{const}$. Can one do
$$frac{partial f}{partial (Cx)}=frac{1}{C}frac{partial f}{partial x}$$
Is this always correct? Or do we need the derivative $f_x'$ to exist, or be continuous before we can write the above?
multivariable-calculus derivatives partial-derivative
asked Dec 18 '18 at 15:56
user142523user142523
18112
$endgroup$
Suppose $f=f(x,y)$ is any function of 2 variables and $C=mbox{const}$. Can one do
$$frac{partial f}{partial (Cx)}=frac{1}{C}frac{partial f}{partial x}$$
Is this always correct? Or do we need the derivative $f_x'$ to exist, or be continuous before we can write the above?
multivariable-calculus derivatives partial-derivative
multivariable-calculus derivatives partial-derivative
asked Dec 18 '18 at 15:56
user142523user142523
18112
asked Dec 18 '18 at 15:56
user142523user142523
18112
edited Dec 18 '18 at 21:52
user142523
asked Dec 18 '18 at 15:56
user142523user142523
18112
asked Dec 18 '18 at 15:56
user142523user142523
18112
asked Dec 18 '18 at 15:56
user142523user142523
18112
18112
add a comment |
add a comment |
1 Answer
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If by "constant" you mean $C$ depends on neither $x$ nor $y$, then for $Cne 0$ you only need $partial f/partial x$ to exist. Necessity is trivial; sufficiency is an exercise. (Hint: use $lim_{xto a}kg(x)=klim_{xto a}g(x)$.)
answered Dec 18 '18 at 16:02
J.G.J.G.
30.6k23149
$endgroup$
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
1
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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$begingroup$
If by "constant" you mean $C$ depends on neither $x$ nor $y$, then for $Cne 0$ you only need $partial f/partial x$ to exist. Necessity is trivial; sufficiency is an exercise. (Hint: use $lim_{xto a}kg(x)=klim_{xto a}g(x)$.)
answered Dec 18 '18 at 16:02
J.G.J.G.
30.6k23149
$endgroup$
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
1
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
add a comment |
$begingroup$
If by "constant" you mean $C$ depends on neither $x$ nor $y$, then for $Cne 0$ you only need $partial f/partial x$ to exist. Necessity is trivial; sufficiency is an exercise. (Hint: use $lim_{xto a}kg(x)=klim_{xto a}g(x)$.)
answered Dec 18 '18 at 16:02
J.G.J.G.
30.6k23149
$endgroup$
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
1
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
add a comment |
$begingroup$
If by "constant" you mean $C$ depends on neither $x$ nor $y$, then for $Cne 0$ you only need $partial f/partial x$ to exist. Necessity is trivial; sufficiency is an exercise. (Hint: use $lim_{xto a}kg(x)=klim_{xto a}g(x)$.)
answered Dec 18 '18 at 16:02
J.G.J.G.
30.6k23149
$endgroup$
If by "constant" you mean $C$ depends on neither $x$ nor $y$, then for $Cne 0$ you only need $partial f/partial x$ to exist. Necessity is trivial; sufficiency is an exercise. (Hint: use $lim_{xto a}kg(x)=klim_{xto a}g(x)$.)
answered Dec 18 '18 at 16:02
J.G.J.G.
30.6k23149
answered Dec 18 '18 at 16:02
J.G.J.G.
30.6k23149
answered Dec 18 '18 at 16:02
J.G.J.G.
30.6k23149
answered Dec 18 '18 at 16:02
J.G.J.G.
30.6k23149
30.6k23149
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
1
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
add a comment |
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
1
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
$begingroup$
Thank you for your answer! What about $frac{partial^2 f}{partial (Cx)partial (Cy)} = frac{1}{C^2}frac{partial^2 f}{partial x partial y}$
$endgroup$
– user142523
Dec 18 '18 at 16:13
1
1
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
$begingroup$
@user142553 Again, we only need the (in this case second) derivative to exist.
$endgroup$
– J.G.
Dec 18 '18 at 16:14
add a comment |
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