# Partial Differentiation -- If w=x+y and s=(x^3)+xy+(y^3), find πw/πs

## Homework Statement:

If w=x+y and s=(x^3)+xy+(y^3), find πw/πs

## Relevant Equations:

πw/πs=(πw/πx)*(πx/πs)+(πw/πy)*(πy/πs)
πw/πx=1

and then I wasn't sure about πx/πs, so I tried implicitly differentiating s:

1=(3x^2)(πx/πs)+y(πx/πs)+x(πy/πs)+(3y^2)(πy/πs)

And then I shaved my head in frustration.

Delta2

Homework Helper
Gold Member
2020 Award
If you are taking a partial derivative, what parameter is being held constant?

haruspex
Homework Helper
Gold Member
2020 Award
Is this the whole question, word for word? It doesn't make sense to me.
If s varies it is because x and or y varies, but a given change in s can be achieved by changes in x and y in different ways, and these lead to different changes in w.

I forgot this part of the question. Apologies: (x^2)*y+x(y^2)=t

Homework Helper
Gold Member
2020 Award
Is ##dt=0 ##? i.e. is the partial derivative w.r.t. ## s ## at fixed ## t ##? You need to be more clear please.

Delta2
This is all I know.

#### Attachments

• 8.9 KB Views: 31
haruspex
Homework Helper
Gold Member
2020 Award
This is all I know.
It could have been made clearer in the question, but from the full text one can guess that the partial wrt s means t constant, and mutatis mutandis.

I apologize for that. Nothing else has flummoxed in calc, but whatever reason, this has done just that.

haruspex
Homework Helper
Gold Member
2020 Award
I apologize for that. Nothing else has flummoxed in calc, but whatever reason, this has done just that.
Are you ok to go ahead now?

Not really.

haruspex
Homework Helper
Gold Member
2020 Award
Not really.
What equation do you get if you ##\partial/\partial s## the equation for t?

Delta2
Homework Helper
Gold Member
Guys i dont understand why you interfere with ##t##. You can evaluate ##\frac{\partial x}{\partial s}## directly as $$\frac{\partial x}{\partial s}=\frac{1}{\frac{\partial s}{\partial x}}$$ you ll just find it as function of x and y instead of s and t.
Same logic applies to ##\frac{\partial y}{\partial s}##.

t is just t(x,y). I don't see how πt/πs help me.

And frankly I don't see how to calculate πt/πs

haruspex
Homework Helper
Gold Member
2020 Award
Guys i dont understand why you interfere with ##t##. You can evaluate ##\frac{\partial x}{\partial s}## directly as $$\frac{\partial x}{\partial s}=\frac{1}{\frac{\partial s}{\partial x}}$$ you ll just find it as function of x and y instead of s and t.
Same logic applies to ##\frac{\partial y}{\partial s}##.
You can't do that with partials.
x=r cos(ΞΈ)
##\partial x/\partial r=\cos(\theta)##
##r^2=x^2+y^2##
##\partial r/\partial x=2x##
The flaw is that something different is being held constant in the two cases.

Homework Helper
Gold Member
2020 Award
What equation do you get if you ##\partial/\partial s## the equation for t?
alternatively (for the OP) you can set ## dt=0 ##.

Delta2
Homework Helper
Gold Member
You can't do that with partials.
x=r cos(ΞΈ)
##\partial x/\partial r=\cos(\theta)##
##r^2=x^2+y^2##
##\partial r/\partial x=2x##
The flaw is that something different is being held constant in the two cases.
Yes you can. You have a mistake in the last line, you get that $$2r\frac{\partial r}{\partial x}=2x\Rightarrow\frac{\partial r}{\partial x}=\frac{x}{r}=\cos\theta$$

Delta2
Homework Helper
Gold Member
Oh sorry , should have prove that ##\frac{\partial r}{\partial x}=\frac{1}{\cos\theta}##, maybe you are right @haruspex .

alternatively (for the OP) you can set ## dt=0 ##.
When you say let dt=0, let the partial of t with respect to x plus the partial of t with respect to y=0?

haruspex
Homework Helper
Gold Member
2020 Award
Yes you can. You have a mistake in the last line, you get that $$2r\frac{\partial r}{\partial x}=2x\Rightarrow\frac{\partial r}{\partial x}=\frac{x}{r}=\cos\theta$$
Yes, that was careless, but as you saw the corrected version makes my point.

alternatively (for the OP) you can set ## dt=0 ##.
So if dt=0 then:

(2xy+y^2)dx+(2xy+x^2)dy=0.....?

haruspex
Homework Helper
Gold Member
2020 Award
When you say let dt=0, let the partial of t with respect to x plus the partial of t with respect to y=0?
No, @Charles Link and I are saying the same thing in different ways. Differentiate the expression for t partially wrt s.
The key is that if there are two independent variables u and v then ##\frac{\partial f}{\partial u}## is shorthand for ##\frac{\partial f}{\partial u}|_{v}##, i.e. the change in f as u varies but v stays constant. In particular, ##\frac{\partial v}{\partial u}=0## by definition.

In my Cartesian/polar example, the partial derivative wrt x implicitly means holding y constant, whereas the partial wrt r means holding theta constant.

t isn't a function of s.

haruspex
Homework Helper
Gold Member
2020 Award
t isn't a function of s.
No matter.. ##\partial t/\partial s## means the change in t if s changes but t doesn't, i.e. 0.
It produces what you got in post #21 except that, usefully, you have ##\partial x/\partial s## in place of dx and ##\partial y/\partial s## in place of dy.

Ok, I see that.