Having problems with the (I think) general chain rule

In summary, the conversation involved a discussion about a transformation \mathbf{x}=G(\mathbf{u}), where \mathbf{x}=(x_1,x_2,x_3),\:\mathbf{u}=(u_1,u_2,u_3) and \mathbf{x} is a function of \mathbf{u}. The focus was on computing the derivative of the transformation and showing its local invertibility under certain conditions. The Jacobian matrix was mentioned as a tool for calculating the derivative. The individual was also looking for additional resources to aid in their understanding, such as a companion text or online resources like Khan Academy.
  • #1
richyw
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0

Homework Statement



Consider the transformation [itex]\mathbf{x}=G(\mathbf{u}), \text{ where } \mathbf{x}=(x_1,x_2,x_3),\:\mathbf{u}=(u_1,u_2,u_3)[/itex] given by[tex]x_1=u_1+u_3^2[/tex][tex]x_2=u_3-u_1^2[/tex][tex]x_3=u_1+u_2+u_3[/tex]

I need to compute the derivative of this transformation, and then show that the transformation is locally invertible if [itex]u_1u_3>0[/itex]

Homework Equations



unsure really.

The Attempt at a Solution



well x is a function of [itex]x_1,x_2,x_3[/itex] and all the [itex]x_j's[/itex] are functions of [itex]u_1,u_2,u_3[/itex]. So I am dealing with a map [itex]\mathbb{R}^3[/itex] to [itex]\mathbb{R}^3[/itex] right?

sorry I'm really lost on what this question is asking. mostly focusing on the firsrt part right now (the derivative)
 
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  • #2
richyw said:

Homework Statement



Consider the transformation [itex]\mathbf{x}=G(\mathbf{u}), \text{ where } \mathbf{x}=(x_1,x_2,x_3),\:\mathbf{u}=(u_1,u_2,u_3)[/itex] given by[tex]x_1=u_1+u_3^2[/tex][tex]x_2=u_3-u_1^2[/tex][tex]u_1+u_2+u_3[/tex]

I need to compute the derivative of this transformation, and then show that the transformation is locally invertible if [itex]u_1u_3>0[/itex]

Homework Equations



unsure really.

The Attempt at a Solution



well x is a function of [itex]x_1,x_2,x_3[/itex] and all the [itex]x_j's[/itex] are functions of [itex]u_1,u_2,u_3[/itex]. So I am dealing with a map [itex]\mathbb{R}^3[/itex] to [itex]\mathbb{R}^3[/itex] right?

sorry I'm really lost on what this question is asking. mostly focusing on the firsrt part right now (the derivative)

I suppose that third line is supposed to read ##x_3=u_1+u_2+u_3##. Is that what is causing your problems? Can you calculate the Jacobian now?
 
  • #3
that's a typo yes, but not what is causing my problem.
 
  • #4
[tex](x_1,x_2,x_3)=G(u_1,u_2,u_3)[/tex]
[tex]D\mathbf{x}=\left[\begin{matrix} \frac{\partial x_1}{\partial u_1} & \frac{\partial x_1}{\partial u_2} & \frac{\partial x_1}{\partial u_3} \\ \frac{\partial x_2}{\partial u_1} & \frac{\partial x_2}{\partial u_2} & \frac{\partial x_2}{\partial x_3} \\ \frac{\partial x_3}{\partial u_1} & \frac{\partial x_3}{\partial u_2} & \frac{\partial x_3}{\partial u_3}\end{matrix}\right]
[/tex]
 
  • #5
is that the jacobian?
 
  • #6
richyw said:
[tex](x_1,x_2,x_3)=G(u_1,u_2,u_3)[/tex]
[tex]D\mathbf{x}=\left[\begin{matrix} \frac{\partial x_1}{\partial u_1} & \frac{\partial x_1}{\partial u_2} & \frac{\partial x_1}{\partial u_3} \\ \frac{\partial x_2}{\partial u_1} & \frac{\partial x_2}{\partial u_2} & \frac{\partial x_2}{\partial x_3} \\ \frac{\partial x_3}{\partial u_1} & \frac{\partial x_3}{\partial u_2} & \frac{\partial x_3}{\partial u_3}\end{matrix}\right]
[/tex]

richyw said:
is that the jacobian?

Yes. You can read about the Jacobian and its properties here:

http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
 
  • #7
ok cool, thanks a lot!

I'm sure i'll be able to hack through the rest of it. I'm quite frustrated with my textbook right now (folland). It seems to be more of a reference text than something that I can actually learn from. Still looking for a companion text, or even better something like khan academy.
 

FAQ: Having problems with the (I think) general chain rule

What is the general chain rule?

The general chain rule is a rule in calculus that allows us to find the derivative of a composite function. It is used when we have a function within another function, and we need to find the derivative of the outer function with respect to the inner function.

When do we use the general chain rule?

The general chain rule is used when we have a function that is composed of two or more functions, and we need to find the derivative of the composite function. It is also used when we have a function that is composed of a function and its inverse.

How do we apply the general chain rule?

To apply the general chain rule, we first identify the inner function and the outer function. Then, we take the derivative of the outer function with respect to the inner function, and multiply it by the derivative of the inner function with respect to the independent variable. This gives us the derivative of the composite function.

What are some common mistakes when using the general chain rule?

One common mistake when using the general chain rule is forgetting to take the derivative of the inner function with respect to the independent variable. Another mistake is taking the derivative of the outer function with respect to the wrong variable. It is important to carefully identify and differentiate between the inner and outer functions to avoid these mistakes.

Can the general chain rule be applied to any composite function?

Yes, the general chain rule can be applied to any composite function, as long as the derivatives of the inner and outer functions exist. However, it is important to note that some composite functions may require more complex rules, such as the product or quotient rule, in addition to the general chain rule.

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