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

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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.