Solving Partial Derivation Homework Problem

[Bman900]

Now I solved something similar to this problem yesterday (https://www.physicsforums.com/showthread.php?t=447168) thanks to the help of p21bass but this one is really out there and I have no idea where to begin.

Homework Statement

Homework Equations

The Attempt at a Solution

I don't know where to even start as this is my first time ever seeing this problem. Where should I start?

 

[robcowlam]

So is the question just asking you to find Fx, Fy, Fz?
If that is the case then its a fairly easy problem.
If you have not met partial differentiation before I won't go through what it is and how it comes about but i'll just tell you how to do it:
To differentiate du/dx (with curly d's to represent partial derivatives) you take all the variables which are not x and treat them as constants so for example d/dx (x^2yz)=2xyz
As with the minus signs you just need to find the derivatives and then multiply it by -1 .
Hopefully that helps a little.

 

[Bman900]

So is the question just asking you to find Fx, Fy, Fz?
If that is the case then its a fairly easy problem.
If you have not met partial differentiation before I won't go through what it is and how it comes about but i'll just tell you how to do it:
To differentiate du/dx (with curly d's to represent partial derivatives) you take all the variables which are not x and treat them as constants so for example d/dx (x^2yz)=2xyz
As with the minus signs you just need to find the derivatives and then multiply it by -1 .
Hopefully that helps a little.

so like this?

But since am treating yz as constants wouldn't it be 0 if I take the derivative or am just taking the derivative of x and then multiplying it by yz?

 

[Bman900]

Ok so I read up on partial derivatives and came up with this:

Am I right?

 

[alex3]

Not quite. When you partially differentiate, you're treating the other variables as constant, but you still might be multiplying by the variable you're differentiating with respect to. For instance:

$$\frac{\partial }{\partial x} ( xyz ) = yz$$

As you know

$$\frac{d}{dx} ( \alpha x ) = \alpha$$

Remember: when you differentiate a constant on its own, you get 0, but a constant multiplying the variable you're differentiating with respect to is not zero!

 

[Bman900]

I really do appreciate the help here! Now is this any better?

 

[alex3]

Looks great, nice work!

 

[robcowlam]

Spot on buddy, hopefully you understand partial derivatives now if not try reading:
http://tutorial.math.lamar.edu/Classes/CalcIII/PartialDerivsIntro.aspx its a really useful maths website.