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

[icez]

Homework Statement

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

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.

 

[xepma]

Let me rewrite it for you:

w^2 + w(xz-yx)+ (yz-5) = 0

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

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

 

[icez]

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:

(yz)x + (z)xw - (yz)w + w^2 - 5 = 0

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

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

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

Thanks again.

 

[HallsofIvy]

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

 

[icez]

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: F(x,y,z,w)=xyz+xzw-yzw+w^2-5=0

\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}

Thank you all for your help.