Question

Use the Implicit Function Theorem to find

Answer 1

\[dz/dy, given \sqrt{3y} -ycos(xz)=2x-3z^2\]

Answer 2

Its partial z partial y. noy dz/dy. Don't know how to write the partial partial thing

Answer 3

\partial is the fancy d symbol :)

Answer 4

is the formula not \[dz/dy= -fy/fz\]

Answer 5

is it? so differentiating w.r.t.y you keep x and z constant

Answer 6

which gives \[3/2\sqrt{3y} - \cos(xz)\]

Answer 7

and w.r.t.z you keep x and y constant

which gives \[xysin(xz) +6z\]

Answer 8

Is it?

Answer 9

\[-[3/\sqrt{3y}-\cos(xz)]/{xysin(xz)+6z}\]

Answer 10

Final answer?

Answer 11

@Loser66 @Michele_Laino

Answer 12

I don't know :)

Answer 13

KK thanks

Answer 14

I think that we have to apply the implicit functions theorem in the case of more than one dimension

Answer 15

Is this more than one dimension?

Answer 16

we have to define a function \(F\) such that:
\[\large F\left( {x,y,z} \right) = \sqrt {3y} - y\cos \left( {xz} \right) - 2x + 3{z^2}\]

Answer 17

Let me try once :)
z = f(x,y) , hence \(\dfrac{\partial f}{\partial y}\) means x is a constant.
We have 4terms.
Now the first term \((\sqrt 3\sqrt y)'= \sqrt 3 *y^{-2/3}\)

Answer 18

of course, we have:
\[\large F\left( {x,y,z} \right) = \sqrt {3y} - y\cos \left( {xz} \right) - 2x + 3{z^2}=0\]

Answer 19

Ok makes sense.

Answer 20

Do the same with other terms. :)

Answer 21

Wrap up: Find dz/dy, given \sqrt{3y} -ycos(xz)=2x-3z^2

by which you mean\[dz/dy,or z _{y}\]

Answer 22

@mathmale yayyy!!! its partial z partial y

Answer 23

the second term: \((-ycos(xz))' = -cos(xz) +ysin(xz)*x\dfrac{dz}{dy}\)

Answer 24

given that given that\[given.that. \sqrt{3y} -ycos(xz)=2x-3z^2\]

Answer 25

the third term =0
the last term = \(-6z\dfrac{dz}{dy}\)
put them on the original equation
isolate \(dz/dy\), then you get the answer

Answer 26

@loser66: Would we be better off using the partial differential operator

Answer 27

instead of that apostrophe?

(-ycos(xz))' = -cos(xz) +ysin(xz)*x\dfrac{dz}{dy}

Answer 28

in more than one dimension, the final thesis, is:
\[\large\frac{{\partial z}}{{\partial \left( {x,y} \right)}} = - {\left( {\frac{{\partial F}}{{\partial z}}\left( {x,y,z\left( {x,y} \right)} \right)} \right)^{ - 1}}\left( {\frac{{\partial F}}{{\partial \left( {x,y} \right)}}\left( {x,y,z\left( {x,y} \right)} \right)} \right)\]

Answer 29

I just "try once" . Not now whether it is right or wrong. :) rely on you guys to check it out.

Answer 30

*know

Answer 31

is loser on the right track, because as @Michele_Laino pointed out f(x y z)

Answer 32

where, by definition, we have:
\[\Large\frac{{\partial z}}{{\partial \left( {x,y} \right)}} \equiv \left( {\frac{{\partial z}}{{\partial x}},\frac{{\partial z}}{{\partial y}}} \right)\]

Answer 33

kk @Loser66