Partial differentiation chain rule, differential operator?
We are given the function
begin{equation}
V(x,y)= f(s)+g(t)
end{equation}
with s=x+y and t=x+0.5y.
How can I calculate $V_{xx}$ and $V_{yy}$?
I have calculated $V_{x}$ and $V_{y}$ but I do not know how to apply the differential operator to the derivatives.
begin{equation}
frac{partial{V}}{partial{x}}=frac{partial{V}}{partial{s}} frac{partial{s}}{partial{x}}+frac{partial{V}}{partial{t}} frac{partial{t}}{partial{x}}=frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}}
end{equation}
Hence,
begin{equation}
V_{xx}=frac{partial{}V_{x}}{partial{x}}=frac{partial{}}{partial{x}}(frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}})=(frac{partial{}}{partial{s}} +frac{partial{}}{partial{t}})(frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}})
end{equation}
where I have plugged in the differential operator. Then, however I would also get mixed derivatives.
begin{equation}
(frac{partial{}}{partial{s}} +frac{partial{}}{partial{t}})(frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}})=V_{ss}+V_{st}+V_{ts}+V_{tt}
end{equation}
Assuming this method is correct, how do I get rid of the mixed derivatives?
partial-derivative chain-rule
asked Dec 28 '18 at 18:39
MrDerDartMrDerDart
124
add a comment |
We are given the function
begin{equation}
V(x,y)= f(s)+g(t)
end{equation}
with s=x+y and t=x+0.5y.
How can I calculate $V_{xx}$ and $V_{yy}$?
I have calculated $V_{x}$ and $V_{y}$ but I do not know how to apply the differential operator to the derivatives.
begin{equation}
frac{partial{V}}{partial{x}}=frac{partial{V}}{partial{s}} frac{partial{s}}{partial{x}}+frac{partial{V}}{partial{t}} frac{partial{t}}{partial{x}}=frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}}
end{equation}
Hence,
begin{equation}
V_{xx}=frac{partial{}V_{x}}{partial{x}}=frac{partial{}}{partial{x}}(frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}})=(frac{partial{}}{partial{s}} +frac{partial{}}{partial{t}})(frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}})
end{equation}
where I have plugged in the differential operator. Then, however I would also get mixed derivatives.
begin{equation}
(frac{partial{}}{partial{s}} +frac{partial{}}{partial{t}})(frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}})=V_{ss}+V_{st}+V_{ts}+V_{tt}
end{equation}
Assuming this method is correct, how do I get rid of the mixed derivatives?
partial-derivative chain-rule
asked Dec 28 '18 at 18:39
MrDerDartMrDerDart
124
add a comment |
We are given the function
begin{equation}
V(x,y)= f(s)+g(t)
end{equation}
with s=x+y and t=x+0.5y.
How can I calculate $V_{xx}$ and $V_{yy}$?
I have calculated $V_{x}$ and $V_{y}$ but I do not know how to apply the differential operator to the derivatives.
begin{equation}
frac{partial{V}}{partial{x}}=frac{partial{V}}{partial{s}} frac{partial{s}}{partial{x}}+frac{partial{V}}{partial{t}} frac{partial{t}}{partial{x}}=frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}}
end{equation}
Hence,
begin{equation}
V_{xx}=frac{partial{}V_{x}}{partial{x}}=frac{partial{}}{partial{x}}(frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}})=(frac{partial{}}{partial{s}} +frac{partial{}}{partial{t}})(frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}})
end{equation}
where I have plugged in the differential operator. Then, however I would also get mixed derivatives.
begin{equation}
(frac{partial{}}{partial{s}} +frac{partial{}}{partial{t}})(frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}})=V_{ss}+V_{st}+V_{ts}+V_{tt}
end{equation}
Assuming this method is correct, how do I get rid of the mixed derivatives?
partial-derivative chain-rule
asked Dec 28 '18 at 18:39
MrDerDartMrDerDart
124
We are given the function
begin{equation}
V(x,y)= f(s)+g(t)
end{equation}
with s=x+y and t=x+0.5y.
How can I calculate $V_{xx}$ and $V_{yy}$?
I have calculated $V_{x}$ and $V_{y}$ but I do not know how to apply the differential operator to the derivatives.
begin{equation}
frac{partial{V}}{partial{x}}=frac{partial{V}}{partial{s}} frac{partial{s}}{partial{x}}+frac{partial{V}}{partial{t}} frac{partial{t}}{partial{x}}=frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}}
end{equation}
Hence,
begin{equation}
V_{xx}=frac{partial{}V_{x}}{partial{x}}=frac{partial{}}{partial{x}}(frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}})=(frac{partial{}}{partial{s}} +frac{partial{}}{partial{t}})(frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}})
end{equation}
where I have plugged in the differential operator. Then, however I would also get mixed derivatives.
begin{equation}
(frac{partial{}}{partial{s}} +frac{partial{}}{partial{t}})(frac{partial{V}}{partial{s}} +frac{partial{V}}{partial{t}})=V_{ss}+V_{st}+V_{ts}+V_{tt}
end{equation}
Assuming this method is correct, how do I get rid of the mixed derivatives?
partial-derivative chain-rule
partial-derivative chain-rule
asked Dec 28 '18 at 18:39
MrDerDartMrDerDart
124
asked Dec 28 '18 at 18:39
MrDerDartMrDerDart
124
edited Dec 29 '18 at 10:56
MrDerDart
asked Dec 28 '18 at 18:39
MrDerDartMrDerDart
124
asked Dec 28 '18 at 18:39
MrDerDartMrDerDart
124
asked Dec 28 '18 at 18:39
MrDerDartMrDerDart
124
124
add a comment |
add a comment |
2 Answers
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We have $V(x,y)=f(x+y)+g(x+0.5y)$ so that
$$V_y(x,y)=f'(x+y)+0.5g'(x+0.5y)$$
and
$$V_{yy}(x,y)=f''(x+y)+0.25g''(x+0.5y)$$
Finding $V_{xx}$ is even easier.
answered Dec 28 '18 at 18:45
Ben WBen W
2,007615
add a comment |
$$V(x,y)= f(s)+g(t)$$
Then $$V_x=frac{partial s}{partial x}f'(s)+frac{partial t}{partial x}g'(t)$$
And so on. This is how you get factors of $1/2$ each time you differentiate $g$ by $y$.
If, say $f$ had been a more complicated function, $f=f(s,t)$, then you'd have $$V_x=frac{partial s}{partial x}frac{partial f}{partial s}+frac{partial t}{partial x}frac{partial f}{partial t}+frac{partial t}{partial x}g'(t)$$
answered Dec 28 '18 at 18:56
John DoeJohn Doe
10.9k11238
I've updated my question.
– MrDerDart
Dec 29 '18 at 10:57
What you have done is correct. You get rid of the mixed derivatives simply due to the fact that neither f nor g are dependent on both s and t. So if you differentiate them both by s and t, they will both go to zero.
– John Doe
Dec 29 '18 at 13:43
add a comment |
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2 Answers
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2 Answers
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We have $V(x,y)=f(x+y)+g(x+0.5y)$ so that
$$V_y(x,y)=f'(x+y)+0.5g'(x+0.5y)$$
and
$$V_{yy}(x,y)=f''(x+y)+0.25g''(x+0.5y)$$
Finding $V_{xx}$ is even easier.
answered Dec 28 '18 at 18:45
Ben WBen W
2,007615
add a comment |
We have $V(x,y)=f(x+y)+g(x+0.5y)$ so that
$$V_y(x,y)=f'(x+y)+0.5g'(x+0.5y)$$
and
$$V_{yy}(x,y)=f''(x+y)+0.25g''(x+0.5y)$$
Finding $V_{xx}$ is even easier.
answered Dec 28 '18 at 18:45
Ben WBen W
2,007615
add a comment |
We have $V(x,y)=f(x+y)+g(x+0.5y)$ so that
$$V_y(x,y)=f'(x+y)+0.5g'(x+0.5y)$$
and
$$V_{yy}(x,y)=f''(x+y)+0.25g''(x+0.5y)$$
Finding $V_{xx}$ is even easier.
answered Dec 28 '18 at 18:45
Ben WBen W
2,007615
We have $V(x,y)=f(x+y)+g(x+0.5y)$ so that
$$V_y(x,y)=f'(x+y)+0.5g'(x+0.5y)$$
and
$$V_{yy}(x,y)=f''(x+y)+0.25g''(x+0.5y)$$
Finding $V_{xx}$ is even easier.
answered Dec 28 '18 at 18:45
Ben WBen W
2,007615
answered Dec 28 '18 at 18:45
Ben WBen W
2,007615
answered Dec 28 '18 at 18:45
Ben WBen W
2,007615
answered Dec 28 '18 at 18:45
Ben WBen W
2,007615
2,007615
add a comment |
add a comment |
$$V(x,y)= f(s)+g(t)$$
Then $$V_x=frac{partial s}{partial x}f'(s)+frac{partial t}{partial x}g'(t)$$
And so on. This is how you get factors of $1/2$ each time you differentiate $g$ by $y$.
If, say $f$ had been a more complicated function, $f=f(s,t)$, then you'd have $$V_x=frac{partial s}{partial x}frac{partial f}{partial s}+frac{partial t}{partial x}frac{partial f}{partial t}+frac{partial t}{partial x}g'(t)$$
answered Dec 28 '18 at 18:56
John DoeJohn Doe
10.9k11238
I've updated my question.
– MrDerDart
Dec 29 '18 at 10:57
What you have done is correct. You get rid of the mixed derivatives simply due to the fact that neither f nor g are dependent on both s and t. So if you differentiate them both by s and t, they will both go to zero.
– John Doe
Dec 29 '18 at 13:43
add a comment |
$$V(x,y)= f(s)+g(t)$$
Then $$V_x=frac{partial s}{partial x}f'(s)+frac{partial t}{partial x}g'(t)$$
And so on. This is how you get factors of $1/2$ each time you differentiate $g$ by $y$.
If, say $f$ had been a more complicated function, $f=f(s,t)$, then you'd have $$V_x=frac{partial s}{partial x}frac{partial f}{partial s}+frac{partial t}{partial x}frac{partial f}{partial t}+frac{partial t}{partial x}g'(t)$$
answered Dec 28 '18 at 18:56
John DoeJohn Doe
10.9k11238
I've updated my question.
– MrDerDart
Dec 29 '18 at 10:57
What you have done is correct. You get rid of the mixed derivatives simply due to the fact that neither f nor g are dependent on both s and t. So if you differentiate them both by s and t, they will both go to zero.
– John Doe
Dec 29 '18 at 13:43
add a comment |
$$V(x,y)= f(s)+g(t)$$
Then $$V_x=frac{partial s}{partial x}f'(s)+frac{partial t}{partial x}g'(t)$$
And so on. This is how you get factors of $1/2$ each time you differentiate $g$ by $y$.
If, say $f$ had been a more complicated function, $f=f(s,t)$, then you'd have $$V_x=frac{partial s}{partial x}frac{partial f}{partial s}+frac{partial t}{partial x}frac{partial f}{partial t}+frac{partial t}{partial x}g'(t)$$
answered Dec 28 '18 at 18:56
John DoeJohn Doe
10.9k11238
$$V(x,y)= f(s)+g(t)$$
Then $$V_x=frac{partial s}{partial x}f'(s)+frac{partial t}{partial x}g'(t)$$
And so on. This is how you get factors of $1/2$ each time you differentiate $g$ by $y$.
If, say $f$ had been a more complicated function, $f=f(s,t)$, then you'd have $$V_x=frac{partial s}{partial x}frac{partial f}{partial s}+frac{partial t}{partial x}frac{partial f}{partial t}+frac{partial t}{partial x}g'(t)$$
answered Dec 28 '18 at 18:56
John DoeJohn Doe
10.9k11238
answered Dec 28 '18 at 18:56
John DoeJohn Doe
10.9k11238
answered Dec 28 '18 at 18:56
John DoeJohn Doe
10.9k11238
answered Dec 28 '18 at 18:56
John DoeJohn Doe
10.9k11238
10.9k11238
I've updated my question.
– MrDerDart
Dec 29 '18 at 10:57
What you have done is correct. You get rid of the mixed derivatives simply due to the fact that neither f nor g are dependent on both s and t. So if you differentiate them both by s and t, they will both go to zero.
– John Doe
Dec 29 '18 at 13:43
add a comment |
I've updated my question.
– MrDerDart
Dec 29 '18 at 10:57
What you have done is correct. You get rid of the mixed derivatives simply due to the fact that neither f nor g are dependent on both s and t. So if you differentiate them both by s and t, they will both go to zero.
– John Doe
Dec 29 '18 at 13:43
I've updated my question.
– MrDerDart
Dec 29 '18 at 10:57
I've updated my question.
– MrDerDart
Dec 29 '18 at 10:57
What you have done is correct. You get rid of the mixed derivatives simply due to the fact that neither f nor g are dependent on both s and t. So if you differentiate them both by s and t, they will both go to zero.
– John Doe
Dec 29 '18 at 13:43
What you have done is correct. You get rid of the mixed derivatives simply due to the fact that neither f nor g are dependent on both s and t. So if you differentiate them both by s and t, they will both go to zero.
– John Doe
Dec 29 '18 at 13:43
add a comment |
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