Partial derivatives of implicitly defined functions

[compliant]

Homework Statement

If the equations

$$x^2 - 2(y^2)(s^2)t - 2st^2 = 1$$
$$x^2 + 2(y^2)(s^2)t + 5st^2 = 1$$

define s and t as functions of x and y, find $$\partial^2 t / \partial y^2$$

The Attempt at a Solution

Equating the two, we get 4y^2*s^2*t = -7s*t^2. My main problem is, as simple as this sounds, how do I implicitly differentiate a single term with three variables? (i.e. 4y^2*s^2*t) Because if I can implicitly differentiate it in terms of y, I can probably find $$\partial t / \partial y$$ in terms of the other partial derivatives.

 

[djeitnstine]

Partial differentiation is not implicit differentiation if that's what you mean. You simply treat all other variables as constant. In this case you get 8y*s^2 etc...

 

[compliant]

But how would I get the partial derivative of t wrt y from there? I mean, aren't we assuming t is a function of x and y (and with s as well?) ?So let me try this out:

8y*s^2*t + 4y^2($$2 \partial s / \partial y$$)t + 4y^2*s^2*($$\partial t / \partial y$$) = -7($$\partial s / \partial y$$)*t^2 - 7s($$2 \partial t / \partial y$$)

 

[gabbagabbahey]

But how would I get the partial derivative of t wrt y from there? I mean, aren't we assuming t is a function of x and y (and with s as well?) ?

Don't combine your two original equations; instead differentiate each one with respect to $y$--- you will end up with two equations, each involving $\frac{\partial s}{\partial y}$ and $\frac{\partial t}{\partial y}$, which you can then solve for $\frac{\partial t}{\partial y}$.

 

[compliant]

Don't combine your two original equations; instead differentiate each one with respect to $y$--- you will end up with two equations, each involving $\frac{\partial s}{\partial y}$ and $\frac{\partial t}{\partial y}$, which you can then solve for $\frac{\partial t}{\partial y}$.

Sounds like a plan. So let me try this again...[-4y$s^2$t - 2$y^2$($2s\frac{\partial s}{\partial y})t$ - 2$y^2 s^2$($\frac{\partial t}{\partial y}$)] + [-2($\frac{\partial s}{\partial y})t^2$ - 2s($2 \frac{\partial t}{\partial y})$] = 1

[4y$s^2$t + 2$y^2$($2s\frac{\partial s}{\partial y})t$ + 2$y^2 s^2$($\frac{\partial t}{\partial y}$)] + [5($\frac{\partial s}{\partial y})t^2$ + 5s($2 \frac{\partial t}{\partial y})$] = 1[3($\frac{\partial s}{\partial y})t^2$ + 3s($2 \frac{\partial t}{\partial y})$] = 2

Is there a way to eliminate the $\frac{\partial s}{\partial y}$ term?

 

[gabbagabbahey]

Sounds like a plan. So let me try this again...


[-4y$s^2$t - 2$y^2$($2s\frac{\partial s}{\partial y})t$ - 2$y^2 s^2$($\frac{\partial t}{\partial y}$)] + [-2($\frac{\partial s}{\partial y})t^2$ - 2s($2 \frac{\partial t}{\partial y})$] = 1

[4y$s^2$t + 2$y^2$($2s\frac{\partial s}{\partial y})t$ + 2$y^2 s^2$($\frac{\partial t}{\partial y}$)] + [5($\frac{\partial s}{\partial y})t^2$ + 5s($2 \frac{\partial t}{\partial y})$] = 1

Careful, when you differentiate an equation, you need to differentiate both sides of it.

$$\frac{\partial}{\partial y} (1)\neq 1$$

Is there a way to eliminate the $\frac{\partial s}{\partial y}$ term?

Sure...you have 2 equations and 2 unkowns; think back to your high school algebra

 

[compliant]

Careful, when you differentiate an equation, you need to differentiate both sides of it.

$$\frac{\partial}{\partial y} (1)\neq 1$$



Sure...you have 2 equations and 2 unkowns; think back to your high school algebra

Ok, so now I have

$$3t^2 \frac {\partial s}{\partial y} = -6s \frac{\partial t}{\partial y}$$
$$\frac {\partial s}{\partial y} = \frac{-2s}{t^2} \frac{\partial t}{\partial y}$$

I sub this back into the first equations to get $\frac{\partial t}{\partial y}$. However, I get a different value for $\frac{\partial t}{\partial y}$ when I plug it into the second equation. If I'm going to write the answer, would I write them both down? Because I feel there really should only be one answer...