How Do You Apply the Chain Rule in Multivariable Calculus Problems?

In summary, the student is attempting to solve a homework problem using a method that he is not familiar with. He is computing the partial derivatives of a matrix of components against a single variable.
  • #1
richyw
180
0

Homework Statement



Let [itex](u,v)=\mathbf{f}(x,y,x)=(2x+y^3,xe^{5y-7z})[/itex]

Compute [itex]D\mathbf{f}(x,y,z),\;\partial (u,v)/\partial(x,y),\;\partial (u,v)/\partial(y,z)\text{ and }\partial (u,v)/\partial(x,z)[/itex]

Homework Equations



-chain rule

The Attempt at a Solution



well I can get [tex]D\mathbf{f}(x,y,z)=\left[\begin{matrix} \partial u / \partial x & \partial u / \partial y & \partial u / \partial z \\ \partial v / \partial x & \partial v / \partial y & \partial v/ \partial z \end{matrix}\right]=\left[\begin{matrix} 2 & 3y^2 & 0 \\ e^{5y-7z} & 5xe^{5y-7z} & -7xe^{5y-7z}\end{matrix}\right][/tex]and I can see that if I ignore the variable that is not in the partial, and take the determinant of that 2x2 matrix, then I get the partial derivative I want. My questions is what am I doing? what happens if it is a 3x2 matrix instead of a 2x3? does this "method" still work?
 
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  • #2
I don't understand the question. You seemed to have solved the problem. (The solution looks good to me). What determinant are you talking about?
 
  • #3
richyw said:
if I ignore the variable that is not in the partial, and take the determinant of that 2x2 matrix, then I get the partial derivative I want.

I do not understand what you mean.
 
  • #4
well I have the matrix derivative, but I still need to get those three partial derivatives. For example I can see that [tex]\frac{\partial (u,v)}{\partial (x,y)}=\left|\begin{matrix} 2 & 3y^2 \\ e^{5y-7z} & 5xe^{5y-7z} \end{matrix}\right|=(10x-3y^2)e^{5y-7z}[/tex]so I am taking the components of my matrix derivative that are in question and taking the determinant of the 2x2 matrix. I just don't get why I am doing this. I don't understand what I'm doing, only how to do it.
 
  • #5
I'm not familiar with the notation ##\partial (u,v)/\partial(x,z)##. How is it defined? I can see how you're computing it, but is it defined as that determinant?
 
  • #6
Fredrik said:
I'm not familiar with the notation ##\partial (u,v)/\partial(x,z)##. How is it defined? I can see how you're computing it, but is it defined as that determinant?

##\frac{∂(x_1,...,x_n)}{∂(v_1,...,v_n)} =

\left| \begin{array}{ccc}
({x_{1}})_{v_1} & ... & ({x_{1}})_{v_n} \\
... & ... & ... \\
({x_{n}})_{v_1} & ... & ({x_{n}})_{v_n} \end{array} \right|##

It's the Jacobian.
 
  • #7
richyw said:
well I have the matrix derivative, but I still need to get those three partial derivatives. For example I can see that [tex]\frac{\partial (u,v)}{\partial (x,y)}=\left|\begin{matrix} 2 & 3y^2 \\ e^{5y-7z} & 5xe^{5y-7z} \end{matrix}\right|=(10x-3y^2)e^{5y-7z}[/tex]so I am taking the components of my matrix derivative that are in question and taking the determinant of the 2x2 matrix. I just don't get why I am doing this. I don't understand what I'm doing, only how to do it.

If you are supposed to find Df, which is a matrix of the partial derivatives of all the components against all the variables, which happens to be 2x3 in this case, then why do you care about a 2x2 matrix (and its determinant) of some partial derivatives? And the answer is, you should not. Just compute all the partials, as requested, and be done with that.
 
  • #8
Zondrina said:
##\frac{∂(x_1,...,x_n)}{∂(v_1,...,v_n)} =

\left| \begin{array}{ccc}
({x_{1}})_{v_1} & ... & ({x_{1}})_{v_n} \\
... & ... & ... \\
({x_{n}})_{v_1} & ... & ({x_{n}})_{v_n} \end{array} \right|##

It's the Jacobian.
Is it really? The u and v that are given in the problem are functions of three variables, but only two of them appear in these "Jacobians".
 
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  • #9
voko said:
If you are supposed to find Df, which is a matrix of the partial derivatives of all the components against all the variables, which happens to be 2x3 in this case, then why do you care about a 2x2 matrix (and its determinant) of some partial derivatives? And the answer is, you should not. Just compute all the partials, as requested, and be done with that.
Apparently he's not just supposed to find Df(x,y,z). The problem is asking specifically for those determinants.
 
  • #10
voko said:
then why do you care about a 2x2 matrix (and its determinant) of some partial derivatives? And the answer is, you should not. Just compute all the partials, as requested, and be done with that.

but that's the thing. I'm getting the 3 requested partials by taking the determinant of those 2x2 matrices. I computed the matrix derivative as requested, but I am not sure what I am actually "doing" when I am asked to find those partial derivatives. I'm not even sure what the thing I am finding represents. All I know how to do is compute it for the test!

my textbook is "Folland, Advanced Calculus" and this is chapter 2.10, i case anyone has that book
 
  • #11
richyw said:
but that's the thing. I'm getting the 3 requested partials by taking the determinant of those 2x2 matrices.
But the requested things aren't partial derivatives, are they? Can you post the definitions and theorems that make you think that you should compute those determinants. (I'm not saying that you're wrong, only that I need to see some definitions).

richyw said:
my textbook is "Folland, Advanced Calculus" and this is chapter 2.10, i case anyone has that book
There's no preview at google books, and I was unable to find a pdf online.
 
  • #12
hmm ok well I think I will have to put this one on hold until the weekend. afterall I can get the marks on the test and that's what school all about right! haha.
 
  • #13
It seems to me that you know how to get the result, but you don't really understand the significance of the result. Is that correct? I am not familiar with that textbook, and the comments on Amazon are not very encouraging, but perhaps you should try re-reading the relevant part on theory.
 

FAQ: How Do You Apply the Chain Rule in Multivariable Calculus Problems?

What is the chain rule?

The chain rule is a rule in calculus that helps us find the derivative of a composite function. It allows us to break down a complex function into simpler functions and calculate the derivative of each part separately.

Why is the chain rule important?

The chain rule is important because it is the fundamental rule used to differentiate composite functions, which are common in many areas of mathematics and science. It allows us to find the rate of change of a function with respect to its independent variables.

How do I know when to use the chain rule?

The chain rule is used when finding the derivative of a composite function, where one function is nested inside another. You can identify a composite function by looking for functions within functions or by noticing that the independent variable appears in multiple places in the function.

Can you provide an example of using the chain rule?

Sure, an example of using the chain rule would be finding the derivative of f(x) = (x^2 + 1)^5. We can break this down into two parts: g(x) = x^2 + 1 and h(x) = x^5. Using the chain rule, we can find the derivative of g(x) and h(x) separately, and then multiply them together to find the derivative of the original function, f(x).

Are there any common mistakes when using the chain rule?

Yes, some common mistakes when using the chain rule include forgetting to apply the derivative to the outer function, not taking the derivative of the inner function, and not simplifying the final answer. It is important to carefully follow the steps of the chain rule to avoid these mistakes.

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