Question

Find the Jacobian formula for
f(x,y)= sin (x^2+y^3)
Please, help. Formula only

Answer 1

@Empty

Answer 2

For some reason, I don't know how to write it down :(
calculating is easy. :)

Answer 3

Can I do: let u = sin x,
\[\dfrac{\partial f}{\partial x}=\dfrac{\partial }{\partial u}(sin (x^2+y^2))\dfrac{\partial (x^2+y^3)}{\partial x} \]
\[\dfrac{\partial f}{\partial y}=\dfrac{\partial }{\partial u}(sin (x^2+y^2))\dfrac{\partial (x^2+y^3)}{\partial y} \]

Answer 4

Isn't there another function, like g(x,y) that goes with this?

Answer 5

Nope, that is all I have. ha!! I have a bunch of them. I need know how to jot the formula down.

Answer 6

\[\frac{∂(x,y,z)}{∂(u,v,w)} =\det \left[\begin{matrix}\frac{∂x}{∂u} & \frac{∂y}{∂u} & \frac{∂z}{∂u} \\ \frac{∂x}{∂v} & \frac{∂y}{∂v} & \frac{∂z}{∂v}\\ \frac {∂x}{∂w} & \frac {∂z}{∂w} & \frac {∂z}{∂w}\end{matrix}\right]\]

\[dxdydz = \left| \frac{∂(x,y,z)}{∂(u,v,w)} \right|dudvdw\]

Answer 7

Jacobians are generally used to change coordinates, so this doesn't really make any sense, unless you're shifting from 2 dimensions onto 1 dimension which is kind of uncommon. Generally you have

f(x,y) and g(x,y) which is the change from x and y to f and g while you can also look at x(f,g) and y(f,g) as two Jacobian matrices. The cool thing about these is that when you take all the partials and put them into their own matrices, they are matrix inverses of each other.

Answer 8

So specifically: \[\left[\begin{matrix}f_x & f_y \\ g_x & g_y\end{matrix}\right]\left[\begin{matrix}x_f & x_g\\ y_f & y_g\end{matrix}\right] = \left[\begin{matrix}1 & 0\\ 0 & 1\end{matrix}\right]\]

Answer 9

This problem is \(f:\mathbb R^2\rightarrow \mathbb R\). It's qualify to find the Jacobian transformation.

Answer 10

\(f: (x,y)\mapsto sin (x^2+y^2)\)

Answer 11

*y^3

Answer 12

I think you already have your answer, here, why isn't this enough?
\[\dfrac{\partial f}{\partial x}=\dfrac{\partial }{\partial u}(sin (x^2+y^2))\dfrac{\partial (x^2+y^3)}{\partial x} \]
\[\dfrac{\partial f}{\partial y}=\dfrac{\partial }{\partial u}(sin (x^2+y^2))\dfrac{\partial (x^2+y^3)}{\partial y} \]

Answer 13

One more question: what is difference between Derivative matrix and Jacobian ?

Answer 14

Nothing, they are the same thing

Answer 15

So for your case you have a 2x1 or 1x2 matrix of partial derivatives since you're going from 2 to 1 dimension.

Answer 16

Another one: :)
what is difference between \(\dfrac{\partial }{\partial u}(sin (x^2+y^2))\)and \(\dfrac{\partial (sin (x^2+y^3))}{\partial x}\)

Answer 17

Well partial u doesn't really make sense, it's just sort of like the chain rule. I see your mistake now I was tired and didn't pay attention but what you should have written to help you out with the chain rule is this: \[\large \text{let } u = x^2+y^2\]\[\large \frac{\partial}{\partial x} \sin(x^2+y^2)=\frac{\partial}{\partial x} \sin(u) \\ \large \cos(u)* \frac{\partial u}{\partial x}\]

Now looking at what u is above we can evaluate \[\large \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(x^2+y^2)=2x\] So we plug it in and get \[\large \frac{\partial f}{\partial x} = 2x \cos(x^2+y^2)\]

Answer 18

I guess I put the wrong power of y there, but it really doesn't matter much here haha.

Answer 19

Thanks for the help. I got it. :)

Answer 20

Ok =)