Partial of F(x,y,z,w) w.r.t x : Solution

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Homework Help Overview

The discussion revolves around finding the partial derivative of w with respect to x for the implicit function defined by F(x,y,z,w) = xyz + xzw - yzw + w² - 5 = 0. Participants are exploring the challenges posed by the presence of w in multiple terms of the equation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the difficulty of isolating w due to its presence in various terms and consider implicit differentiation as a potential method. Some attempt to rearrange the equation to facilitate differentiation, while others express uncertainty about their results.

Discussion Status

Several participants have shared their attempts at implicit differentiation, with varying degrees of confidence in their results. There is a mix of approaches being explored, and while some guidance has been offered, no consensus has been reached on the correct method or outcome.

Contextual Notes

Participants note the complexity introduced by the w² term and the implications of treating y and z as constants during differentiation. There is an acknowledgment of potential confusion regarding the application of implicit differentiation in this context.

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Homework Statement



Find [tex]\frac{\partial{w}}{\partial{x}}[/tex] for [tex]F(x,y,z,w)=xyz+xzw-yzw+w^2-5=0[/tex]

The Attempt at a Solution


I know how to get partials (such as Dz/Dx) of functions in the form z = f(x,y). I'm having trouble figuring out this one since it's asking for the partial of w, which is all over the function. I tried putting everything equal to w, but that w2 doesn't help. I thought of perhaps doing implicit differentiation, but I still end up with that extra w on one side.

I'm sure it's something tiny that I'm not seeing, but I need a little push in the right direction.

Thank you much.
 
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Let me rewrite it for you:

[tex]w^2 + w(xz-yx)+ (yz-5) = 0[/tex]

Now if I also write this:
[tex]w^2 + a*w + b = 0[/tex]

Does it ring a bell what the next step should be?
 


I tried that, but that would seem to only get everything factored, or to find the values of w when F(x,y,z) is 0. I could be wrong though. I tried something else, I figured since y and z are constants I rearranged it as:

[tex](yz)x + (z)xw - (yz)w + w^2 - 5 = 0[/tex]

Then I just took the partials of all that, and got:

[tex]\frac{\partial{w}}{\partial{x}} = 1(yz) + (x+w)(z) - 1(yz) + 2w = 2w + zw + zx[/tex]

I have no way to check if this is right, but hopefully it is.

Thanks again.
 


Use "implicit differentiation". For example, if [itex]F(x,y,w)= xw+ yw+ w^2= 0[/itex], assuming that w is a function of x and y but x and y are independent variables, then
[tex]\frac{\partial F}{\partial x}= w+ x\frac{\partial w}{\partial x}+ y\frac{\partial w}{\partial x}+ 2w\frac{\partial w}{\partial x}= 0[/tex]
so
[tex](x+ y+ 2w)\frac{\partial w}{\partial x}= -w[/tex]
and
[tex]\frac{\partial w}{\partial x}= \frac{-w}{x+ y + 2w}[/tex]
 


I actually had tried implicit differentiation...I don't know why I thought it was wrong because of the w2. Anyways, I tried it again:

Problem: [tex]F(x,y,z,w)=xyz+xzw-yzw+w^2-5=0[/tex]

[tex]\frac{\partial{w}}{\partial{x}}=0+w+x\frac{\partial{w}}{\partial{x}}-yz\frac{\partial{w}}{\partial{x}}+2w\frac{\partial{w}}{\partial{x}}= \frac{-w}{x-yz+2w}[/tex]

Thank you all for your help.
 

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