# Partial derivative using differentials

1. Aug 1, 2014

1. The problem statement, all variables and given/known data
If $xs^2 + yt^2 = 1$ and $x^2s + y^2t = xy - 4$, find $\frac{\partial x}{\partial s}, \frac{\partial x}{\partial t}, \frac{\partial y}{\partial s}, \frac{\partial y}{\partial t}$, at (x, y, s, t) = (1, -3, 2, -1)

2. Relevant equations

3. The attempt at a solution
I took the differential of both equations and substituted in the values and I got:

$$4ds + 4dx + 6dt + dy = 0$$ and
$$ds + 7dx + 9dt + 5dy = 0$$

but now I don't know how do find the individual partial derivatives that are being asked for. For example, for $\frac{\partial x}{\partial s}$, I assumed t and y are constant and made dt and dy = 0 and tried to solve for ds and dx but that doesn't work.

Any ideas on how to proceed in finding these partials?
Thanks!

2. Aug 1, 2014

$s$ and $t$ don't happen to be independent, so that $\frac{\partial s}{\partial t} = \frac{\partial t}{\partial s} = 0$? That would make things much easier.

3. Aug 1, 2014

### HallsofIvy

Staff Emeritus
In order that this problem make any sense, s and t have to be independent so Quesadilla's comment is crucial. You should NOT assume y is constant since it is one of the dependent variables. But, taking the derivatives with respect to s you should assume t is constant (I would have said "treat t as a constant").

So from $4ds+ 4dx+ 6dt+ dy= 0$ and $ds+ 7dx+ 9dt+ 5dy= 0$ you can get
$$4\frac{\partial s}{\partial s}+ 4\frac{\partial x}{\partial s}+ 6\frac{\partial t}{\partial s}+ \frac{\partial y}{\partial s}= 4+ 4\frac{\partial x}{\partial s}+ \frac{\partial y}{\partial s}= 0$$
and
$$\frac{\partial s}{\partial s}+ 7\frac{\partial x}{\partial s}+ 9\frac{\partial t}{\partial s}+ 5\frac{\partial y}{\partial s}= 1+ 7\frac{\partial x}{\partial s}+ 5\frac{\partial y}{\partial s}= 0$$
Giving you two equations to solve for $\partial x/\partial s$ and $\partial y/\partial$.

Do the same thing to find the derivatives with respect to t.

Personally, I wouldn't have use "differentials". Just differentiating $xs^2+ yt^2= 1$ with respect to s (and treating t as a constant) gives
$$s^2\frac{\partial x}{\partial s}+ 2xs+ t^2\frac{\partial y}{\partial s}= 0$$
and differentiating $x^2s+ y^2t= xy- 4$ with respect to s gives
$$2sx\frac{\partial x}{\partial s}+ x^2+ 2ty\frac{\partial y}{\partial s}= x\frac{\partial y}{\partial s}+ y\frac{\partial x}{\partial s}$$
again giving two equations to solve for $\partial x/\partial s$ and $\partial y/\partial s$.

Last edited: Aug 1, 2014
4. Aug 1, 2014

### Ray Vickson

Why do you claim are s and t not independent? If we arbitrarily specify values of s and t we can then use the equations to find x and y. In fact, I would say we can take s and t as independent, but with x and y being functions of them: x = x(s,t) and y = y(s,t).

Last edited: Aug 1, 2014
5. Aug 1, 2014