Partial Differential
$begingroup$
If f is a function of x. Some times people write:
Is this a rule? does it have a proof? need help.
Thanks.
derivatives
asked Dec 13 '18 at 9:22
DiamondDiamond
143
$endgroup$
add a comment |
$begingroup$
If f is a function of x. Some times people write:
Is this a rule? does it have a proof? need help.
Thanks.
derivatives
asked Dec 13 '18 at 9:22
DiamondDiamond
143
$endgroup$
add a comment |
$begingroup$
If f is a function of x. Some times people write:
Is this a rule? does it have a proof? need help.
Thanks.
derivatives
asked Dec 13 '18 at 9:22
DiamondDiamond
143
$endgroup$
If f is a function of x. Some times people write:
Is this a rule? does it have a proof? need help.
Thanks.
derivatives
derivatives
asked Dec 13 '18 at 9:22
DiamondDiamond
143
asked Dec 13 '18 at 9:22
DiamondDiamond
143
asked Dec 13 '18 at 9:22
DiamondDiamond
143
asked Dec 13 '18 at 9:22
DiamondDiamond
143
asked Dec 13 '18 at 9:22
DiamondDiamond
143
143
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
begin{align*}
x_i &= x_i(t) text{ and}\
F &= f(x_1, x_2, ..., x_n)\
frac{dF}{dt} &= sum^n_{i=1} frac{partial f}{partial x_i} frac{dx_i}{dt}\
dF &= sum^n_{i=1} frac{partial f}{partial x_i} dx_i
end{align*}
In your case, F is a function of x, let x be the $x_1$, that is, $F = f(x_1)$, so
begin{align*}
dF &=frac{partial f}{partial x_1} dx_1 \
&= frac{df}{dx_1} dx_1
end{align*}
answered Dec 13 '18 at 9:41
TEDTED
34
$endgroup$
$begingroup$
OK but why capital F?
$endgroup$
– Diamond
Dec 13 '18 at 17:40
$begingroup$
That's just an arbitrary symbol, notice that F=f anyway.
$endgroup$
– TED
Dec 14 '18 at 1:29
add a comment |
$begingroup$
$df=frac{partial f}{partial x}dx$ is false if $f$ is a function of several variables (which is suggested by your question about partial derivatives). For example in case of $f(x,y)$ : $$df=frac{partial f}{partial x}dx+frac{partial f}{partial y}dy$$ If $f$ is function of one variable only, the "partial derivative" is no longer partial but is the usual derivative $$df=f'(x)dx=left(frac{text{d}f}{text{d}x}right)dx$$ $left(frac{text{d}f}{text{d}x}right)$ denotes a function, not a fraction.
In this case of one variable only, sometimes $frac{partial f}{partial x}$ is loosely confused with $frac{text{d}f}{text{d}x}$. That is a matter of conventional symbols.
answered Dec 13 '18 at 9:45
JJacquelinJJacquelin
44.2k21853
$endgroup$
$begingroup$
so you mean although they wrote it as partial derivative they mean common derivative.
$endgroup$
– Diamond
Dec 13 '18 at 15:32
$begingroup$
"common" : "usual" or "ordinary" (not quite sure of the correct translation).
$endgroup$
– JJacquelin
Dec 13 '18 at 15:51
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
begin{align*}
x_i &= x_i(t) text{ and}\
F &= f(x_1, x_2, ..., x_n)\
frac{dF}{dt} &= sum^n_{i=1} frac{partial f}{partial x_i} frac{dx_i}{dt}\
dF &= sum^n_{i=1} frac{partial f}{partial x_i} dx_i
end{align*}
In your case, F is a function of x, let x be the $x_1$, that is, $F = f(x_1)$, so
begin{align*}
dF &=frac{partial f}{partial x_1} dx_1 \
&= frac{df}{dx_1} dx_1
end{align*}
answered Dec 13 '18 at 9:41
TEDTED
34
$endgroup$
$begingroup$
OK but why capital F?
$endgroup$
– Diamond
Dec 13 '18 at 17:40
$begingroup$
That's just an arbitrary symbol, notice that F=f anyway.
$endgroup$
– TED
Dec 14 '18 at 1:29
add a comment |
$begingroup$
begin{align*}
x_i &= x_i(t) text{ and}\
F &= f(x_1, x_2, ..., x_n)\
frac{dF}{dt} &= sum^n_{i=1} frac{partial f}{partial x_i} frac{dx_i}{dt}\
dF &= sum^n_{i=1} frac{partial f}{partial x_i} dx_i
end{align*}
In your case, F is a function of x, let x be the $x_1$, that is, $F = f(x_1)$, so
begin{align*}
dF &=frac{partial f}{partial x_1} dx_1 \
&= frac{df}{dx_1} dx_1
end{align*}
answered Dec 13 '18 at 9:41
TEDTED
34
$endgroup$
$begingroup$
OK but why capital F?
$endgroup$
– Diamond
Dec 13 '18 at 17:40
$begingroup$
That's just an arbitrary symbol, notice that F=f anyway.
$endgroup$
– TED
Dec 14 '18 at 1:29
add a comment |
$begingroup$
begin{align*}
x_i &= x_i(t) text{ and}\
F &= f(x_1, x_2, ..., x_n)\
frac{dF}{dt} &= sum^n_{i=1} frac{partial f}{partial x_i} frac{dx_i}{dt}\
dF &= sum^n_{i=1} frac{partial f}{partial x_i} dx_i
end{align*}
In your case, F is a function of x, let x be the $x_1$, that is, $F = f(x_1)$, so
begin{align*}
dF &=frac{partial f}{partial x_1} dx_1 \
&= frac{df}{dx_1} dx_1
end{align*}
answered Dec 13 '18 at 9:41
TEDTED
34
$endgroup$
begin{align*}
x_i &= x_i(t) text{ and}\
F &= f(x_1, x_2, ..., x_n)\
frac{dF}{dt} &= sum^n_{i=1} frac{partial f}{partial x_i} frac{dx_i}{dt}\
dF &= sum^n_{i=1} frac{partial f}{partial x_i} dx_i
end{align*}
In your case, F is a function of x, let x be the $x_1$, that is, $F = f(x_1)$, so
begin{align*}
dF &=frac{partial f}{partial x_1} dx_1 \
&= frac{df}{dx_1} dx_1
end{align*}
answered Dec 13 '18 at 9:41
TEDTED
34
edited Dec 13 '18 at 9:46
answered Dec 13 '18 at 9:41
TEDTED
34
answered Dec 13 '18 at 9:41
TEDTED
34
answered Dec 13 '18 at 9:41
TEDTED
34
34
$begingroup$
OK but why capital F?
$endgroup$
– Diamond
Dec 13 '18 at 17:40
$begingroup$
That's just an arbitrary symbol, notice that F=f anyway.
$endgroup$
– TED
Dec 14 '18 at 1:29
add a comment |
$begingroup$
OK but why capital F?
$endgroup$
– Diamond
Dec 13 '18 at 17:40
$begingroup$
That's just an arbitrary symbol, notice that F=f anyway.
$endgroup$
– TED
Dec 14 '18 at 1:29
$begingroup$
OK but why capital F?
$endgroup$
– Diamond
Dec 13 '18 at 17:40
$begingroup$
OK but why capital F?
$endgroup$
– Diamond
Dec 13 '18 at 17:40
$begingroup$
That's just an arbitrary symbol, notice that F=f anyway.
$endgroup$
– TED
Dec 14 '18 at 1:29
$begingroup$
That's just an arbitrary symbol, notice that F=f anyway.
$endgroup$
– TED
Dec 14 '18 at 1:29
add a comment |
$begingroup$
$df=frac{partial f}{partial x}dx$ is false if $f$ is a function of several variables (which is suggested by your question about partial derivatives). For example in case of $f(x,y)$ : $$df=frac{partial f}{partial x}dx+frac{partial f}{partial y}dy$$ If $f$ is function of one variable only, the "partial derivative" is no longer partial but is the usual derivative $$df=f'(x)dx=left(frac{text{d}f}{text{d}x}right)dx$$ $left(frac{text{d}f}{text{d}x}right)$ denotes a function, not a fraction.
In this case of one variable only, sometimes $frac{partial f}{partial x}$ is loosely confused with $frac{text{d}f}{text{d}x}$. That is a matter of conventional symbols.
answered Dec 13 '18 at 9:45
JJacquelinJJacquelin
44.2k21853
$endgroup$
$begingroup$
so you mean although they wrote it as partial derivative they mean common derivative.
$endgroup$
– Diamond
Dec 13 '18 at 15:32
$begingroup$
"common" : "usual" or "ordinary" (not quite sure of the correct translation).
$endgroup$
– JJacquelin
Dec 13 '18 at 15:51
add a comment |
$begingroup$
$df=frac{partial f}{partial x}dx$ is false if $f$ is a function of several variables (which is suggested by your question about partial derivatives). For example in case of $f(x,y)$ : $$df=frac{partial f}{partial x}dx+frac{partial f}{partial y}dy$$ If $f$ is function of one variable only, the "partial derivative" is no longer partial but is the usual derivative $$df=f'(x)dx=left(frac{text{d}f}{text{d}x}right)dx$$ $left(frac{text{d}f}{text{d}x}right)$ denotes a function, not a fraction.
In this case of one variable only, sometimes $frac{partial f}{partial x}$ is loosely confused with $frac{text{d}f}{text{d}x}$. That is a matter of conventional symbols.
answered Dec 13 '18 at 9:45
JJacquelinJJacquelin
44.2k21853
$endgroup$
$begingroup$
so you mean although they wrote it as partial derivative they mean common derivative.
$endgroup$
– Diamond
Dec 13 '18 at 15:32
$begingroup$
"common" : "usual" or "ordinary" (not quite sure of the correct translation).
$endgroup$
– JJacquelin
Dec 13 '18 at 15:51
add a comment |
$begingroup$
$df=frac{partial f}{partial x}dx$ is false if $f$ is a function of several variables (which is suggested by your question about partial derivatives). For example in case of $f(x,y)$ : $$df=frac{partial f}{partial x}dx+frac{partial f}{partial y}dy$$ If $f$ is function of one variable only, the "partial derivative" is no longer partial but is the usual derivative $$df=f'(x)dx=left(frac{text{d}f}{text{d}x}right)dx$$ $left(frac{text{d}f}{text{d}x}right)$ denotes a function, not a fraction.
In this case of one variable only, sometimes $frac{partial f}{partial x}$ is loosely confused with $frac{text{d}f}{text{d}x}$. That is a matter of conventional symbols.
answered Dec 13 '18 at 9:45
JJacquelinJJacquelin
44.2k21853
$endgroup$
$df=frac{partial f}{partial x}dx$ is false if $f$ is a function of several variables (which is suggested by your question about partial derivatives). For example in case of $f(x,y)$ : $$df=frac{partial f}{partial x}dx+frac{partial f}{partial y}dy$$ If $f$ is function of one variable only, the "partial derivative" is no longer partial but is the usual derivative $$df=f'(x)dx=left(frac{text{d}f}{text{d}x}right)dx$$ $left(frac{text{d}f}{text{d}x}right)$ denotes a function, not a fraction.
In this case of one variable only, sometimes $frac{partial f}{partial x}$ is loosely confused with $frac{text{d}f}{text{d}x}$. That is a matter of conventional symbols.
answered Dec 13 '18 at 9:45
JJacquelinJJacquelin
44.2k21853
edited Dec 13 '18 at 15:46
answered Dec 13 '18 at 9:45
JJacquelinJJacquelin
44.2k21853
answered Dec 13 '18 at 9:45
JJacquelinJJacquelin
44.2k21853
answered Dec 13 '18 at 9:45
JJacquelinJJacquelin
44.2k21853
44.2k21853
$begingroup$
so you mean although they wrote it as partial derivative they mean common derivative.
$endgroup$
– Diamond
Dec 13 '18 at 15:32
$begingroup$
"common" : "usual" or "ordinary" (not quite sure of the correct translation).
$endgroup$
– JJacquelin
Dec 13 '18 at 15:51
add a comment |
$begingroup$
so you mean although they wrote it as partial derivative they mean common derivative.
$endgroup$
– Diamond
Dec 13 '18 at 15:32
$begingroup$
"common" : "usual" or "ordinary" (not quite sure of the correct translation).
$endgroup$
– JJacquelin
Dec 13 '18 at 15:51
$begingroup$
so you mean although they wrote it as partial derivative they mean common derivative.
$endgroup$
– Diamond
Dec 13 '18 at 15:32
$begingroup$
so you mean although they wrote it as partial derivative they mean common derivative.
$endgroup$
– Diamond
Dec 13 '18 at 15:32
$begingroup$
"common" : "usual" or "ordinary" (not quite sure of the correct translation).
$endgroup$
– JJacquelin
Dec 13 '18 at 15:51
$begingroup$
"common" : "usual" or "ordinary" (not quite sure of the correct translation).
$endgroup$
– JJacquelin
Dec 13 '18 at 15:51
add a comment |
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