Partial Differentiation -- If w=x+y and s=(x^3)+xy+(y^3), find 𝝏w/𝝏s

[Delta2]

Something just doesn't look quite right to me, I guess because we change on the fly what is supposed to be kept constant, but anyway I guess that's how the chain rule is supposed to work.

 

[Robin64]

Thanks, all, for the help. Using Kramer's rule to solve for the answer made it nice and easy.

 

[Office_Shredder]

I think the polar coordinates example demonstrating you can flip the partial derivatives is misleading because small moves in r and theta are orthogonal to each other, or maybe was just a lucky choice of coordinates.

Let t=x, s=x+y. Then $\partial s / \partial x = 1$, but when you flip the coordinates you get x=t,y=s-t, and $\partial x / \partial s= 0$.

The proper thing to do is to write down the Jacobian matrix, and then when you flip the coordinates you can just invert the Jacobian.

 

[Charles Link]

Ok fine but something doesn't look quite right to me. The chain rule is
$$1=\frac{\partial s}{\partial s}=\frac{\partial s}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial s}{\partial y}\frac{\partial y}{\partial s}$$.
In the above when we calculate for example $\frac{\partial s}{\partial x}$ we keep constant y, but when we calculate $\frac{\partial x}{\partial s}$ we keep constant t. Isn't that contradicting or it's that how the chain rule is supposed to work?

In this problem, it is something of the form $s=u_a v_a+u_b v_b$. In this case $\frac{\partial{s}}{\partial{s}}=u_a (\frac{\partial{v_a}}{\partial{s}})+v_a (\frac{ \partial{u_a}}{\partial{s}}) + (similarly \, for \, b)$. The partials on $u$ and $v$ are done with the chain rule. (We also have $s=s(u_a, v_a, u_b,v_b)$, and we apply the chain rule in taking the partial w.r.t. $s$, e.g. first taking the partial w.r.t. $u_a$, and treating the others as constants, etc., and then taking partial w.r.t. $v_a$, etc.).

 

[Delta2]

In this problem, it is something of the form $s=u_a v_a+u_b v_b$. In this case $\frac{\partial{s}}{\partial{s}}=u_a (\frac{\partial{v_a}}{\partial{s}})+v_a (\frac{ \partial{u_a}}{\partial{s}}) + (similarly \, for \, b)$. The partials on $u$ and $v$ are done with the chain rule. (We also have $s=s(u_a, v_a, u_b,v_b)$, and we apply the chain rule in taking the partial w.r.t. $s$, e.g. first taking the partial w.r.t. $u_a$, and treating the others as constants, etc., and then taking partial w.r.t. $v_a$, etc.).

Hm, sorry I just can't understand what are the $u_a,v_a,u_b,v_b$. Don't we just take $s=x^3+xy+y^3$ and apply the chain rule as in post #34 to get 1 equation and then also do similar $$0=\frac{\partial t}{\partial s}=\frac{\partial t}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial t}{\partial y}\frac{\partial y}{\partial s}$$ to get the other equation with $t=x^2y+xy^2$

 

[Charles Link]

I think it is something we have all done so often that we do it almost automatically. The partials are difficult to write out in Latex, so let me do implicit differentials as the first step: $dt=3x^2 \, dx+x\, dy+y \, dx+3y^2 \, dy$. You then make each differential a partial w.r.t. $s$.

 

[haruspex]

I think the polar coordinates example demonstrating you can flip the partial derivatives is misleading

Which post are you referring to?

 

[Office_Shredder]

Which post are you referring to?

#17, though I now realize #18 is observing that it didn't work, so I guess my example was not helpful.

 

[haruspex]

I did it this way to verify the answer of post 28, and it's not too difficult if you use Kramer's rule type solution to the two linear equations.

Do you mean Cramer's rule?

 

[Charles Link]

Do you mean Cramer's rule?

Did I misspell it? I remember it from my second course in high school algebra, which was about 50 years ago. I googled it now, and yes, I misspelled it.

 

[haruspex]

Did I misspell it? I remember it from my second course in high school algebra, which was about 50 years ago. I googled it now, and yes, I misspelled it.

Ok. A case of Kramer v. Cramer.

I solved it this way:
s looks a bit like ( x+y)³, so try
$s=x^3+xy+y^3=w^3-3x^2y-3xy^2+xy$
$=w^3-3t+xy$
$t=xy(x+y)=xyw$
$xy=\frac tw$
$s=w^3-3t+\frac tw$
Now diff wrt s.

 

[Charles Link]

Ok. A case of Kramer v. Cramer.

In the Dustin Hoffman movie, "Kramer vs. Kramer", it is spelled with a "K". LOL :)

 

[haruspex]

In the Dustin Hoffman movie, "Kramer vs. Kramer", it is spelled with a "K". LOL :)

But what I hadn't noticed is that the movie title misspells versus as "vs.". That's the plural form; it should just be "v."

 

[Delta2]

Ok. A case of Kramer v. Cramer.

I solved it this way:
s looks a bit like ( x+y)³, so try
$s=x^3+xy+y^3=w^3-3x^2y-3xy^2+xy$
$=w^3-3t+xy$
$t=xy(x+y)=xyw$
$xy=\frac tw$
$s=w^3-3t+\frac tw$
Now diff wrt s.

Nice work indeed. With a little bit of algebra you get one equation for s that can give us both partials of w when differentiating with respect to s or t. Very clever shortcut, well done!.

 

[anuttarasammyak]

Thnks to @harupsex #46
$$s=w^3-3t+\frac{t}{w}$$
Differentiating it
$$ds-(3w^2-\frac{t}{w^2})dw+(3-\frac{1}{w})dt=0$$
So we get
$$\frac{\partial w}{\partial s}|_t=\frac{1}{3w^2-\frac{t}{w^2}}$$
$$\frac{\partial w}{\partial t}|_s=\frac{3-\frac{1}{w}}{3w^2-\frac{t}{w^2}}$$