Partial Derivatives

1. Oct 13, 2008

cse63146

Find $$\frac{\partial y}{\partial x}$$ of $$3sin (x^2 + y^2) = 5cos(x^2 - y^2)$$

$$\partial y = 6ycos(x^2 + y^2) - 10ysin(x^2 - y^2)$$
$$\partial x = 6xcos(x^2 + y^2) + 10xsin(x^2 - y^2)$$

so I thought $$\frac{\partial y}{\partial x} = \frac{6ycos(x^2 + y^2) - 10ysin(x^2 - y^2)}{6xcos(x^2 + y^2) + 10xsin(x^2 - y^2)}$$

$$\frac{\partial y}{\partial x} = - \frac{6xcos(x^2 + y^2) + 10xsin(x^2 - y^2)}{6ycos(x^2 + y^2) - 10ysin(x^2 - y^2)}$$

2. Oct 13, 2008

Staff: Mentor

I don't see how you went from an equation to two partial differentials. What happened to the '=' in your first equation?

Instead, take the partial with respect to x of both sides, keeping in mind that the partial of x with respect to x is 1. You'll have an equation that involves x, y, and partial of y w.r.t. x, and you should be able to solve for this partial by using algebra.

3. Oct 13, 2008

cse63146

isnt $$3sin (x^2 + y^2) = 5cos(x^2 - y^2) \Rightarrow 3sin (x^2 + y^2) - 5cos(x^2 - y^2) = 0$$

4. Oct 13, 2008

Staff: Mentor

Yes, the first equation implies the second. In your post, you skipped a lot of steps between where you started and the next line.

If you take the partial w.r.t x of 3 sin(x^2 + y^2) - 5 cos(x^2 - y^2), you're going to have factors of partial wrt x of (x^2 + y^2) and of (x^2 - y^2). For the first, you should get
2x + 2y*partial y w.r.t x. (Sorry I'm not able to use Tex, the database seems very slow and cranky today.) The other factor is similar, but with - between the two terms.
To summarize, after you take partials w.r.t. x of both sides of the equation, you should get this:
3 cos(x^2 + y^2)*(2x + 2y*partial y w.r.t. x) + 5 sin(x^2 - y^2) * (2x + 2y*partial y w.r.t. x) = 0.

You should be able to solve algebraically for partial y w.r.t. x.

An important point is that if you start with an equation, and apply an operator to both sides, you end up with an equation.