Multivariable functions - chain rule

In summary: The conversation involves solving two questions, the first being finding dz/dt for a given equation and the second being showing a proof involving partial derivatives. For the first question, the product rule is used to find dz/dt and then the chain rule is used to find dx/dt. For the second question, the chain rule for partial derivatives is used to find dz/dx and dz/dy. The final step is to substitute these values into the given expression to prove the statement. However, there seems to be a discrepancy in the given question, as the values for u and v may be mixed up.
  • #1
hex.halo
13
0

Homework Statement



Since both my questions are on the same topic, i'll throw them both in here

1. Find dz/dt for z=(x^2)(t^2), x^2+3xt+2t^2=1

2. Show that if u=xy, v=xy and z=f(u,v) then:
x.dz/dx-y.dz/dy=(x-y)dz/dv

Homework Equations





The Attempt at a Solution



1. I only know how to do this when I have something like:

z=x+y, x=t+..., y=t+...

2. I realize that after I get the partial derivatives I just have to substitute back to prove this, but I'm not sure how to get the derivatives. I think it's the z=f(u,v) that has me lost
 
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  • #2
For 1. find dz/dt using the product rule. Then you realize you need dx/dt. Find this by implicitly differentiating the second expression. For the second one review the chain rule for partial derivatives. But it looks to me like the left hand side of the expression you want to prove evaluates to zero. Are you sure you transcribed that right?
 
  • #3
The second question seems wrong!
Is this from Glyn james and that#2 has two parts.. and is like at the bottom of the page on left side column?
One of them should be x+y or x-y
 
  • #4
You're right. It should be u=x+y

This still doesn't help me though. I may be wrong, but I've got:

dz/dx=dz/du.du/dx+dz/dv.dv/dx

dz/dy=dz/du.du/dy+dz/dv.dv/dy

as my partial fractions. But how do I get my dz/du and dz/dv when I have z=f(u,v) and not, for example, z=f(u,v)=2u+3v^2 - I'm lost without this little equation on the end like in previous problems I've done.

Also, I don't think I've understood your explanation of my first question properly. I tried:

dz/dt=dz/dx.dx/dt

dz/dx=2xt^2
dx/dt=(-3x-4t)/(2x+3t)

And therefore:

dz/dt=(-6(x^2)(t^2)-8xt^3)/(2x+3t)

but my textbook's telling me it should be:

4xt(x^2-2t^2)/(2x+3t)
 
  • #5
[tex]\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} \ + \ \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}[/tex]

Let [tex]y=t^2.[/tex]

Hence [tex]\frac{dz}{dt} = 2xt^2 \left( \frac{-3x-4t}{2x+3t} \right) + x^2(2t)[/tex]

Just simplify the above and you're done.

For the 2nd question, is it supposed to be u=xy, v=x+y? I can get the answer if so, but not if the values of u and v are switched as you have put it.
 
  • #6
My sheet's telling me it should be u=x+y, v=xy, but let's just say it's u=xy and v=x+y for the hell of it. How would you go about solving this problem now?
 
  • #7
I don't know if you have copied the question wrongly or if there is something wrong with my working, but here it is:

[tex]\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} \ + \ \frac{\partial z}{\partial v} \frac{\partial v}{\partial x} = \frac{\partial z}{\partial u} + y\frac{\partial z}{\partial v}[/tex]

[tex]\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} \ + \ \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} \ = \ \frac{\partial z}{\partial u} \ + \ x \frac{\partial z}{\partial v}[/tex]

[tex]\frac{\partial u}{\partial x} = 1 \\ \frac{\partial u}{\partial y} = 1[/tex]
[tex]\frac{\partial v}{\partial x} = y \\ \frac{\partial v}{\partial y} = x[/tex]

Hence:

[tex]x\frac{\partial z}{\partial x} - y\frac{\partial z}{\partial y} = (x-y)\frac{\partial z}{\partial u}[/tex]
 

Related to Multivariable functions - chain rule

1. What is the chain rule for multivariable functions?

The chain rule for multivariable functions is a method for finding the derivative of a function with multiple variables. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function.

2. How do you apply the chain rule to multivariable functions?

To apply the chain rule to multivariable functions, you first need to identify the outer and inner functions. Then, take the derivative of the outer function and multiply it by the derivative of the inner function. If there are multiple inner functions, you will need to use the chain rule again for each one.

3. Can the chain rule be extended to more than two variables?

Yes, the chain rule can be extended to functions with more than two variables. The process is the same as for two variables, but the partial derivatives will need to be taken with respect to each variable separately before multiplying them together.

4. How does the chain rule differ from the product rule?

The chain rule is used for finding the derivative of a composite function, while the product rule is used for finding the derivative of a product of two functions. The chain rule involves taking the derivative of the outer function and multiplying it by the derivative of the inner function, while the product rule involves taking the derivative of each function and adding them together.

5. When is the chain rule most commonly used?

The chain rule is most commonly used when dealing with functions that involve multiple variables, such as in multivariable calculus or in physics and engineering problems. It is also used in machine learning and data analysis to find the rate of change of a function with respect to its input variables.

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