Partial derivatives of implicitly defined functions

1. May 30, 2009

compliant

1. The problem statement, all variables and given/known data
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 $$\partial^2 t / \partial y^2$$

3. 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.

2. May 30, 2009

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...

3. May 30, 2009

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($$2 \partial t / \partial y$$)

4. May 30, 2009

gabbagabbahey

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 $\frac{\partial t}{\partial y}$.

5. May 30, 2009

compliant

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?

6. May 30, 2009

gabbagabbahey

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

7. May 31, 2009

compliant

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...

Last edited by a moderator: May 31, 2009