Simplify Derivative with Substitution
up vote
3
down vote
favorite
I try to evaluate:
$$ frac{partial}{partial x} log{u(x, y, z)}$$
Mathematica gives:
$$ frac{1}{x+y+z}$$
I want to simplify the expression with my function:
$$ frac{1}{u(x, y, z)}$$
How to do that?
Thanks.
u[x_, y_, z_] = x + y + z
Simplify[D[Log[u[x, y, z]], x]]
calculus-and-analysis simplifying-expressions
asked Nov 22 at 17:33
R zu
1797
add a comment |
up vote
3
down vote
favorite
I try to evaluate:
$$ frac{partial}{partial x} log{u(x, y, z)}$$
Mathematica gives:
$$ frac{1}{x+y+z}$$
I want to simplify the expression with my function:
$$ frac{1}{u(x, y, z)}$$
How to do that?
Thanks.
u[x_, y_, z_] = x + y + z
Simplify[D[Log[u[x, y, z]], x]]
calculus-and-analysis simplifying-expressions
asked Nov 22 at 17:33
R zu
1797
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I try to evaluate:
$$ frac{partial}{partial x} log{u(x, y, z)}$$
Mathematica gives:
$$ frac{1}{x+y+z}$$
I want to simplify the expression with my function:
$$ frac{1}{u(x, y, z)}$$
How to do that?
Thanks.
u[x_, y_, z_] = x + y + z
Simplify[D[Log[u[x, y, z]], x]]
calculus-and-analysis simplifying-expressions
asked Nov 22 at 17:33
R zu
1797
I try to evaluate:
$$ frac{partial}{partial x} log{u(x, y, z)}$$
Mathematica gives:
$$ frac{1}{x+y+z}$$
I want to simplify the expression with my function:
$$ frac{1}{u(x, y, z)}$$
How to do that?
Thanks.
u[x_, y_, z_] = x + y + z
Simplify[D[Log[u[x, y, z]], x]]
calculus-and-analysis simplifying-expressions
calculus-and-analysis simplifying-expressions
asked Nov 22 at 17:33
R zu
1797
asked Nov 22 at 17:33
R zu
1797
asked Nov 22 at 17:33
R zu
1797
asked Nov 22 at 17:33
R zu
1797
asked Nov 22 at 17:33
R zu
1797
1797
add a comment |
add a comment |
2 Answers
2
active
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votes
up vote
7
down vote
accepted
D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]
1/u[x, y, z]
answered Nov 22 at 17:36
kglr
175k9197402
A more general substitution:/. u[x_,y_,z_] -> Defer[u[x,y,z]]
– R zu
Nov 22 at 18:04
@Rzu, good point.
– kglr
Nov 22 at 18:05
add a comment |
up vote
4
down vote
An alternative is to define UpValues instead of DownValues of u:
Derivative[1, 0, 0][u] ^:= 1&
Derivative[0, 1, 0][u] ^:= 1&
Derivative[0, 0, 1][u] ^:= 1&
D[Log[u[x, y, z]], x]
1/u[x, y, z]
answered Nov 22 at 19:26
Carl Woll
66.7k385174
What are UpValues and DownValues? The definition in the doc seems recursive:UpValue"gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
– R zu
Nov 22 at 19:29
@Rzu Maybe you can check out the documentation forUpSetDelayedandTagSetDelayed.
– Carl Woll
Nov 22 at 19:39
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]
1/u[x, y, z]
answered Nov 22 at 17:36
kglr
175k9197402
A more general substitution:/. u[x_,y_,z_] -> Defer[u[x,y,z]]
– R zu
Nov 22 at 18:04
@Rzu, good point.
– kglr
Nov 22 at 18:05
add a comment |
up vote
7
down vote
accepted
D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]
1/u[x, y, z]
answered Nov 22 at 17:36
kglr
175k9197402
A more general substitution:/. u[x_,y_,z_] -> Defer[u[x,y,z]]
– R zu
Nov 22 at 18:04
@Rzu, good point.
– kglr
Nov 22 at 18:05
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]
1/u[x, y, z]
answered Nov 22 at 17:36
kglr
175k9197402
D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]
1/u[x, y, z]
answered Nov 22 at 17:36
kglr
175k9197402
edited Nov 22 at 18:05
answered Nov 22 at 17:36
kglr
175k9197402
answered Nov 22 at 17:36
kglr
175k9197402
answered Nov 22 at 17:36
kglr
175k9197402
175k9197402
A more general substitution:/. u[x_,y_,z_] -> Defer[u[x,y,z]]
– R zu
Nov 22 at 18:04
@Rzu, good point.
– kglr
Nov 22 at 18:05
add a comment |
A more general substitution:/. u[x_,y_,z_] -> Defer[u[x,y,z]]
– R zu
Nov 22 at 18:04
@Rzu, good point.
– kglr
Nov 22 at 18:05
A more general substitution:
/. u[x_,y_,z_] -> Defer[u[x,y,z]]– R zu
Nov 22 at 18:04
A more general substitution:
/. u[x_,y_,z_] -> Defer[u[x,y,z]]– R zu
Nov 22 at 18:04
@Rzu, good point.
– kglr
Nov 22 at 18:05
@Rzu, good point.
– kglr
Nov 22 at 18:05
add a comment |
up vote
4
down vote
An alternative is to define UpValues instead of DownValues of u:
Derivative[1, 0, 0][u] ^:= 1&
Derivative[0, 1, 0][u] ^:= 1&
Derivative[0, 0, 1][u] ^:= 1&
D[Log[u[x, y, z]], x]
1/u[x, y, z]
answered Nov 22 at 19:26
Carl Woll
66.7k385174
What are UpValues and DownValues? The definition in the doc seems recursive:UpValue"gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
– R zu
Nov 22 at 19:29
@Rzu Maybe you can check out the documentation forUpSetDelayedandTagSetDelayed.
– Carl Woll
Nov 22 at 19:39
add a comment |
up vote
4
down vote
An alternative is to define UpValues instead of DownValues of u:
Derivative[1, 0, 0][u] ^:= 1&
Derivative[0, 1, 0][u] ^:= 1&
Derivative[0, 0, 1][u] ^:= 1&
D[Log[u[x, y, z]], x]
1/u[x, y, z]
answered Nov 22 at 19:26
Carl Woll
66.7k385174
What are UpValues and DownValues? The definition in the doc seems recursive:UpValue"gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
– R zu
Nov 22 at 19:29
@Rzu Maybe you can check out the documentation forUpSetDelayedandTagSetDelayed.
– Carl Woll
Nov 22 at 19:39
add a comment |
up vote
4
down vote
up vote
4
down vote
An alternative is to define UpValues instead of DownValues of u:
Derivative[1, 0, 0][u] ^:= 1&
Derivative[0, 1, 0][u] ^:= 1&
Derivative[0, 0, 1][u] ^:= 1&
D[Log[u[x, y, z]], x]
1/u[x, y, z]
answered Nov 22 at 19:26
Carl Woll
66.7k385174
An alternative is to define UpValues instead of DownValues of u:
Derivative[1, 0, 0][u] ^:= 1&
Derivative[0, 1, 0][u] ^:= 1&
Derivative[0, 0, 1][u] ^:= 1&
D[Log[u[x, y, z]], x]
1/u[x, y, z]
answered Nov 22 at 19:26
Carl Woll
66.7k385174
answered Nov 22 at 19:26
Carl Woll
66.7k385174
answered Nov 22 at 19:26
Carl Woll
66.7k385174
answered Nov 22 at 19:26
Carl Woll
66.7k385174
66.7k385174
What are UpValues and DownValues? The definition in the doc seems recursive:UpValue"gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
– R zu
Nov 22 at 19:29
@Rzu Maybe you can check out the documentation forUpSetDelayedandTagSetDelayed.
– Carl Woll
Nov 22 at 19:39
add a comment |
What are UpValues and DownValues? The definition in the doc seems recursive:UpValue"gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
– R zu
Nov 22 at 19:29
@Rzu Maybe you can check out the documentation forUpSetDelayedandTagSetDelayed.
– Carl Woll
Nov 22 at 19:39
What are UpValues and DownValues? The definition in the doc seems recursive:
UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "– R zu
Nov 22 at 19:29
What are UpValues and DownValues? The definition in the doc seems recursive:
UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "– R zu
Nov 22 at 19:29
@Rzu Maybe you can check out the documentation for
UpSetDelayed and TagSetDelayed.– Carl Woll
Nov 22 at 19:39
@Rzu Maybe you can check out the documentation for
UpSetDelayed and TagSetDelayed.– Carl Woll
Nov 22 at 19:39
add a comment |
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