Finding the partial derivatives of $f(x,y,2x^2+ y^2) = 2x +5y$
$begingroup$
So, given that $f(x,y,2x^2+ y^2) = 2x + 5y$,
I chose $z(x,y) = 2x^2 + y^2$, so that $f(x, y, z(x,y)) = 2x + 5y$.
I just can't figure out the partial derivatives for $x$, $y$, and $z$ of $f$.
What I get -
for $f_x$ -
$f_x + f_z * z_x = 2x$ (I just tried to differentiate the whole equation).
but what is $f_z$?
help :(
derivatives partial-derivative
asked Dec 16 '18 at 18:53
TomTom
132
$endgroup$
add a comment |
$begingroup$
So, given that $f(x,y,2x^2+ y^2) = 2x + 5y$,
I chose $z(x,y) = 2x^2 + y^2$, so that $f(x, y, z(x,y)) = 2x + 5y$.
I just can't figure out the partial derivatives for $x$, $y$, and $z$ of $f$.
What I get -
for $f_x$ -
$f_x + f_z * z_x = 2x$ (I just tried to differentiate the whole equation).
but what is $f_z$?
help :(
derivatives partial-derivative
asked Dec 16 '18 at 18:53
TomTom
132
$endgroup$
1
$begingroup$
Why that third argument in the function?
$endgroup$
– Garmekain
Dec 16 '18 at 19:01
$begingroup$
Your implicit equation only defines $f$ on the surface $z = 2x^2 + y^2$.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 19:10
add a comment |
$begingroup$
So, given that $f(x,y,2x^2+ y^2) = 2x + 5y$,
I chose $z(x,y) = 2x^2 + y^2$, so that $f(x, y, z(x,y)) = 2x + 5y$.
I just can't figure out the partial derivatives for $x$, $y$, and $z$ of $f$.
What I get -
for $f_x$ -
$f_x + f_z * z_x = 2x$ (I just tried to differentiate the whole equation).
but what is $f_z$?
help :(
derivatives partial-derivative
asked Dec 16 '18 at 18:53
TomTom
132
$endgroup$
So, given that $f(x,y,2x^2+ y^2) = 2x + 5y$,
I chose $z(x,y) = 2x^2 + y^2$, so that $f(x, y, z(x,y)) = 2x + 5y$.
I just can't figure out the partial derivatives for $x$, $y$, and $z$ of $f$.
What I get -
for $f_x$ -
$f_x + f_z * z_x = 2x$ (I just tried to differentiate the whole equation).
but what is $f_z$?
help :(
derivatives partial-derivative
derivatives partial-derivative
asked Dec 16 '18 at 18:53
TomTom
132
asked Dec 16 '18 at 18:53
TomTom
132
asked Dec 16 '18 at 18:53
TomTom
132
asked Dec 16 '18 at 18:53
TomTom
132
asked Dec 16 '18 at 18:53
TomTom
132
132
1
$begingroup$
Why that third argument in the function?
$endgroup$
– Garmekain
Dec 16 '18 at 19:01
$begingroup$
Your implicit equation only defines $f$ on the surface $z = 2x^2 + y^2$.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 19:10
add a comment |
1
$begingroup$
Why that third argument in the function?
$endgroup$
– Garmekain
Dec 16 '18 at 19:01
$begingroup$
Your implicit equation only defines $f$ on the surface $z = 2x^2 + y^2$.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 19:10
1
1
$begingroup$
Why that third argument in the function?
$endgroup$
– Garmekain
Dec 16 '18 at 19:01
$begingroup$
Why that third argument in the function?
$endgroup$
– Garmekain
Dec 16 '18 at 19:01
$begingroup$
Your implicit equation only defines $f$ on the surface $z = 2x^2 + y^2$.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 19:10
$begingroup$
Your implicit equation only defines $f$ on the surface $z = 2x^2 + y^2$.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 19:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your implicit equation only defines $f$ on the surface $z = 2x^2 + y^2$. $f_z(x_0,y_0,z_0)$ requires that $f$ be defined at least in a subset of the form ${(x_0,y_0,z):zin(z_0-delta,z_0+delta)}$.
answered Dec 16 '18 at 19:15
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
$endgroup$
$begingroup$
OK, and what if I want to find $D_uf(x_0, y_0, z_0)$? Is it the same gradient as $Z$’s? Because the equation defines $f$ On the surface $z$?
$endgroup$
– Tom
Dec 16 '18 at 19:25
$begingroup$
@Tom, what is $D_u$? A directional derivative? None can be calculated by the same reason.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 20:39
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Your implicit equation only defines $f$ on the surface $z = 2x^2 + y^2$. $f_z(x_0,y_0,z_0)$ requires that $f$ be defined at least in a subset of the form ${(x_0,y_0,z):zin(z_0-delta,z_0+delta)}$.
answered Dec 16 '18 at 19:15
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
$endgroup$
$begingroup$
OK, and what if I want to find $D_uf(x_0, y_0, z_0)$? Is it the same gradient as $Z$’s? Because the equation defines $f$ On the surface $z$?
$endgroup$
– Tom
Dec 16 '18 at 19:25
$begingroup$
@Tom, what is $D_u$? A directional derivative? None can be calculated by the same reason.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 20:39
add a comment |
$begingroup$
Your implicit equation only defines $f$ on the surface $z = 2x^2 + y^2$. $f_z(x_0,y_0,z_0)$ requires that $f$ be defined at least in a subset of the form ${(x_0,y_0,z):zin(z_0-delta,z_0+delta)}$.
answered Dec 16 '18 at 19:15
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
$endgroup$
$begingroup$
OK, and what if I want to find $D_uf(x_0, y_0, z_0)$? Is it the same gradient as $Z$’s? Because the equation defines $f$ On the surface $z$?
$endgroup$
– Tom
Dec 16 '18 at 19:25
$begingroup$
@Tom, what is $D_u$? A directional derivative? None can be calculated by the same reason.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 20:39
add a comment |
$begingroup$
Your implicit equation only defines $f$ on the surface $z = 2x^2 + y^2$. $f_z(x_0,y_0,z_0)$ requires that $f$ be defined at least in a subset of the form ${(x_0,y_0,z):zin(z_0-delta,z_0+delta)}$.
answered Dec 16 '18 at 19:15
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
$endgroup$
Your implicit equation only defines $f$ on the surface $z = 2x^2 + y^2$. $f_z(x_0,y_0,z_0)$ requires that $f$ be defined at least in a subset of the form ${(x_0,y_0,z):zin(z_0-delta,z_0+delta)}$.
answered Dec 16 '18 at 19:15
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
edited Dec 16 '18 at 20:39
answered Dec 16 '18 at 19:15
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
answered Dec 16 '18 at 19:15
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
answered Dec 16 '18 at 19:15
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
34.6k42971
$begingroup$
OK, and what if I want to find $D_uf(x_0, y_0, z_0)$? Is it the same gradient as $Z$’s? Because the equation defines $f$ On the surface $z$?
$endgroup$
– Tom
Dec 16 '18 at 19:25
$begingroup$
@Tom, what is $D_u$? A directional derivative? None can be calculated by the same reason.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 20:39
add a comment |
$begingroup$
OK, and what if I want to find $D_uf(x_0, y_0, z_0)$? Is it the same gradient as $Z$’s? Because the equation defines $f$ On the surface $z$?
$endgroup$
– Tom
Dec 16 '18 at 19:25
$begingroup$
@Tom, what is $D_u$? A directional derivative? None can be calculated by the same reason.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 20:39
$begingroup$
OK, and what if I want to find $D_uf(x_0, y_0, z_0)$? Is it the same gradient as $Z$’s? Because the equation defines $f$ On the surface $z$?
$endgroup$
– Tom
Dec 16 '18 at 19:25
$begingroup$
OK, and what if I want to find $D_uf(x_0, y_0, z_0)$? Is it the same gradient as $Z$’s? Because the equation defines $f$ On the surface $z$?
$endgroup$
– Tom
Dec 16 '18 at 19:25
$begingroup$
@Tom, what is $D_u$? A directional derivative? None can be calculated by the same reason.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 20:39
$begingroup$
@Tom, what is $D_u$? A directional derivative? None can be calculated by the same reason.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 20:39
add a comment |
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1
$begingroup$
Why that third argument in the function?
$endgroup$
– Garmekain
Dec 16 '18 at 19:01
$begingroup$
Your implicit equation only defines $f$ on the surface $z = 2x^2 + y^2$.
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 16 '18 at 19:10