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Chain rule of partial derivatives for composite functions.
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If I have a function of the form $$f(x^2+y^2)$$
How do I find the partial derivatives $$fracpartial fpartial y,fracpartial fpartial x$$
I am not sure how $f(x^2+y^2)$ behaves. I am assuming it should of the form
$$g(x,y)cdot2yquadtextorquad h(x,y)cdot2x$$
calculus partial-derivative
asked Jul 18 at 8:08
Piyush Divyanakar
3,258122
add a comment |Â
up vote
0
down vote
favorite
If I have a function of the form $$f(x^2+y^2)$$
How do I find the partial derivatives $$fracpartial fpartial y,fracpartial fpartial x$$
I am not sure how $f(x^2+y^2)$ behaves. I am assuming it should of the form
$$g(x,y)cdot2yquadtextorquad h(x,y)cdot2x$$
calculus partial-derivative
asked Jul 18 at 8:08
Piyush Divyanakar
3,258122
3
The notation $partial f/partial x$ and $partial f/partial y$ isn't very good, since $f$ is a function of one variable (which hasn't been given a name here). You should give the composite function a name, say $g(x,y)=f(x^2+y^2)$. Then $g$ is a function of two variables and you can compute the partial derivatives $partial g/partial x$ and $partial g/partial y$ in terms of the ordinary one-variable derivative $f'$.
– Hans Lundmark
Jul 18 at 8:44
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If I have a function of the form $$f(x^2+y^2)$$
How do I find the partial derivatives $$fracpartial fpartial y,fracpartial fpartial x$$
I am not sure how $f(x^2+y^2)$ behaves. I am assuming it should of the form
$$g(x,y)cdot2yquadtextorquad h(x,y)cdot2x$$
calculus partial-derivative
asked Jul 18 at 8:08
Piyush Divyanakar
3,258122
If I have a function of the form $$f(x^2+y^2)$$
How do I find the partial derivatives $$fracpartial fpartial y,fracpartial fpartial x$$
I am not sure how $f(x^2+y^2)$ behaves. I am assuming it should of the form
$$g(x,y)cdot2yquadtextorquad h(x,y)cdot2x$$
calculus partial-derivative
asked Jul 18 at 8:08
Piyush Divyanakar
3,258122
edited Jul 18 at 8:14
asked Jul 18 at 8:08
Piyush Divyanakar
3,258122
asked Jul 18 at 8:08
Piyush Divyanakar
3,258122
asked Jul 18 at 8:08
Piyush Divyanakar
3,258122
3,258122
3
The notation $partial f/partial x$ and $partial f/partial y$ isn't very good, since $f$ is a function of one variable (which hasn't been given a name here). You should give the composite function a name, say $g(x,y)=f(x^2+y^2)$. Then $g$ is a function of two variables and you can compute the partial derivatives $partial g/partial x$ and $partial g/partial y$ in terms of the ordinary one-variable derivative $f'$.
– Hans Lundmark
Jul 18 at 8:44
add a comment |Â
3
The notation $partial f/partial x$ and $partial f/partial y$ isn't very good, since $f$ is a function of one variable (which hasn't been given a name here). You should give the composite function a name, say $g(x,y)=f(x^2+y^2)$. Then $g$ is a function of two variables and you can compute the partial derivatives $partial g/partial x$ and $partial g/partial y$ in terms of the ordinary one-variable derivative $f'$.
– Hans Lundmark
Jul 18 at 8:44
3
3
The notation $partial f/partial x$ and $partial f/partial y$ isn't very good, since $f$ is a function of one variable (which hasn't been given a name here). You should give the composite function a name, say $g(x,y)=f(x^2+y^2)$. Then $g$ is a function of two variables and you can compute the partial derivatives $partial g/partial x$ and $partial g/partial y$ in terms of the ordinary one-variable derivative $f'$.
– Hans Lundmark
Jul 18 at 8:44
The notation $partial f/partial x$ and $partial f/partial y$ isn't very good, since $f$ is a function of one variable (which hasn't been given a name here). You should give the composite function a name, say $g(x,y)=f(x^2+y^2)$. Then $g$ is a function of two variables and you can compute the partial derivatives $partial g/partial x$ and $partial g/partial y$ in terms of the ordinary one-variable derivative $f'$.
– Hans Lundmark
Jul 18 at 8:44
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
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accepted
Since $f(.)$ is univariate function we have$$dfracpartial fpartial x=2xf'(x^2+y^2)$$similarly$$dfracpartial fpartial y=2yf'(x^2+y^2)$$therefore the differential is $$df=2xf'(x^2+y^2)dx+2yf'(x^2+y^2)dy$$therefore $$g(x,y)=h(x,y)=f'(x^2+y^2)$$
edited Jul 18 at 8:21
P. Siehr
2,9941722
answered Jul 18 at 8:17
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
1
down vote
First, I want to use some different notation:
$$f(x^2+y^2)=:f(g(x,y))=(fcirc g)(x,y) $$
Now the partial derivatives can be computed by the chain rule (in multiple dimensions):
$$frac∂f∂x = frac∂f∂gfrac∂g∂x = frac∂f∂g2x. $$
$$frac∂f∂y = frac∂f∂gfrac∂g∂y = frac∂f∂g2y. $$
Since you don't know anything else about $f$, you can't simplify these terms further.
answered Jul 18 at 8:19
P. Siehr
2,9941722
The notation $partial f/partial g$ is very misleading in this case, since $f$ is a function of one variable. Further, your $partial f/partial x, partial f/partial y$ should be $partial (f circ g)/partial x, partial (f circ g)/partial y$.
– Gibbs
Jul 18 at 9:39
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since $f(.)$ is univariate function we have$$dfracpartial fpartial x=2xf'(x^2+y^2)$$similarly$$dfracpartial fpartial y=2yf'(x^2+y^2)$$therefore the differential is $$df=2xf'(x^2+y^2)dx+2yf'(x^2+y^2)dy$$therefore $$g(x,y)=h(x,y)=f'(x^2+y^2)$$
edited Jul 18 at 8:21
P. Siehr
2,9941722
answered Jul 18 at 8:17
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
2
down vote
accepted
Since $f(.)$ is univariate function we have$$dfracpartial fpartial x=2xf'(x^2+y^2)$$similarly$$dfracpartial fpartial y=2yf'(x^2+y^2)$$therefore the differential is $$df=2xf'(x^2+y^2)dx+2yf'(x^2+y^2)dy$$therefore $$g(x,y)=h(x,y)=f'(x^2+y^2)$$
edited Jul 18 at 8:21
P. Siehr
2,9941722
answered Jul 18 at 8:17
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since $f(.)$ is univariate function we have$$dfracpartial fpartial x=2xf'(x^2+y^2)$$similarly$$dfracpartial fpartial y=2yf'(x^2+y^2)$$therefore the differential is $$df=2xf'(x^2+y^2)dx+2yf'(x^2+y^2)dy$$therefore $$g(x,y)=h(x,y)=f'(x^2+y^2)$$
edited Jul 18 at 8:21
P. Siehr
2,9941722
answered Jul 18 at 8:17
Mostafa Ayaz
8,6023630
Since $f(.)$ is univariate function we have$$dfracpartial fpartial x=2xf'(x^2+y^2)$$similarly$$dfracpartial fpartial y=2yf'(x^2+y^2)$$therefore the differential is $$df=2xf'(x^2+y^2)dx+2yf'(x^2+y^2)dy$$therefore $$g(x,y)=h(x,y)=f'(x^2+y^2)$$
edited Jul 18 at 8:21
P. Siehr
2,9941722
answered Jul 18 at 8:17
Mostafa Ayaz
8,6023630
edited Jul 18 at 8:21
P. Siehr
2,9941722
edited Jul 18 at 8:21
P. Siehr
2,9941722
edited Jul 18 at 8:21
P. Siehr
2,9941722
2,9941722
answered Jul 18 at 8:17
Mostafa Ayaz
8,6023630
answered Jul 18 at 8:17
Mostafa Ayaz
8,6023630
answered Jul 18 at 8:17
Mostafa Ayaz
8,6023630
8,6023630
add a comment |Â
add a comment |Â
up vote
1
down vote
First, I want to use some different notation:
$$f(x^2+y^2)=:f(g(x,y))=(fcirc g)(x,y) $$
Now the partial derivatives can be computed by the chain rule (in multiple dimensions):
$$frac∂f∂x = frac∂f∂gfrac∂g∂x = frac∂f∂g2x. $$
$$frac∂f∂y = frac∂f∂gfrac∂g∂y = frac∂f∂g2y. $$
Since you don't know anything else about $f$, you can't simplify these terms further.
answered Jul 18 at 8:19
P. Siehr
2,9941722
The notation $partial f/partial g$ is very misleading in this case, since $f$ is a function of one variable. Further, your $partial f/partial x, partial f/partial y$ should be $partial (f circ g)/partial x, partial (f circ g)/partial y$.
– Gibbs
Jul 18 at 9:39
add a comment |Â
up vote
1
down vote
First, I want to use some different notation:
$$f(x^2+y^2)=:f(g(x,y))=(fcirc g)(x,y) $$
Now the partial derivatives can be computed by the chain rule (in multiple dimensions):
$$frac∂f∂x = frac∂f∂gfrac∂g∂x = frac∂f∂g2x. $$
$$frac∂f∂y = frac∂f∂gfrac∂g∂y = frac∂f∂g2y. $$
Since you don't know anything else about $f$, you can't simplify these terms further.
answered Jul 18 at 8:19
P. Siehr
2,9941722
The notation $partial f/partial g$ is very misleading in this case, since $f$ is a function of one variable. Further, your $partial f/partial x, partial f/partial y$ should be $partial (f circ g)/partial x, partial (f circ g)/partial y$.
– Gibbs
Jul 18 at 9:39
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First, I want to use some different notation:
$$f(x^2+y^2)=:f(g(x,y))=(fcirc g)(x,y) $$
Now the partial derivatives can be computed by the chain rule (in multiple dimensions):
$$frac∂f∂x = frac∂f∂gfrac∂g∂x = frac∂f∂g2x. $$
$$frac∂f∂y = frac∂f∂gfrac∂g∂y = frac∂f∂g2y. $$
Since you don't know anything else about $f$, you can't simplify these terms further.
answered Jul 18 at 8:19
P. Siehr
2,9941722
First, I want to use some different notation:
$$f(x^2+y^2)=:f(g(x,y))=(fcirc g)(x,y) $$
Now the partial derivatives can be computed by the chain rule (in multiple dimensions):
$$frac∂f∂x = frac∂f∂gfrac∂g∂x = frac∂f∂g2x. $$
$$frac∂f∂y = frac∂f∂gfrac∂g∂y = frac∂f∂g2y. $$
Since you don't know anything else about $f$, you can't simplify these terms further.
answered Jul 18 at 8:19
P. Siehr
2,9941722
answered Jul 18 at 8:19
P. Siehr
2,9941722
answered Jul 18 at 8:19
P. Siehr
2,9941722
answered Jul 18 at 8:19
P. Siehr
2,9941722
2,9941722
The notation $partial f/partial g$ is very misleading in this case, since $f$ is a function of one variable. Further, your $partial f/partial x, partial f/partial y$ should be $partial (f circ g)/partial x, partial (f circ g)/partial y$.
– Gibbs
Jul 18 at 9:39
add a comment |Â
The notation $partial f/partial g$ is very misleading in this case, since $f$ is a function of one variable. Further, your $partial f/partial x, partial f/partial y$ should be $partial (f circ g)/partial x, partial (f circ g)/partial y$.
– Gibbs
Jul 18 at 9:39
The notation $partial f/partial g$ is very misleading in this case, since $f$ is a function of one variable. Further, your $partial f/partial x, partial f/partial y$ should be $partial (f circ g)/partial x, partial (f circ g)/partial y$.
– Gibbs
Jul 18 at 9:39
The notation $partial f/partial g$ is very misleading in this case, since $f$ is a function of one variable. Further, your $partial f/partial x, partial f/partial y$ should be $partial (f circ g)/partial x, partial (f circ g)/partial y$.
– Gibbs
Jul 18 at 9:39
add a comment |Â
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3
The notation $partial f/partial x$ and $partial f/partial y$ isn't very good, since $f$ is a function of one variable (which hasn't been given a name here). You should give the composite function a name, say $g(x,y)=f(x^2+y^2)$. Then $g$ is a function of two variables and you can compute the partial derivatives $partial g/partial x$ and $partial g/partial y$ in terms of the ordinary one-variable derivative $f'$.
– Hans Lundmark
Jul 18 at 8:44