# Homework Help: Implicit Differentiation in Multivariable Calculus

1. Mar 6, 2006

### Black Orpheus

I need to compute the partials of z with respect to x and y of:
xy + z + 3xz^5 = 4 at (1,0).

I already showed that the equation is solvable for z as a function of (x,y) near (1,0,1) with the special implicit function theorem, but that's the easy part. Could someone explain to me how to begin this?

2. Mar 6, 2006

### Black Orpheus

Do I start by taking d/dx of xy + z + 3xz^4 - 4 = 0, giving
y + dz/dx + 3z^5 = 0, or dz/dx = -y - 3z^5..?

3. Mar 6, 2006

### 0rthodontist

You take d/dx of both sides, taking z as a function of x and y, and taking x and y as not functions of each other (treat y as a constant when taking d/dx, treat x as a constant when taking d/dy). Use the chain rule. For example, d(2x^2 * z^2)/dx = 2xz^2 + 4x^2 * z(dz/dx).

4. Mar 6, 2006

### Black Orpheus

How do I take z as a function of x and y when I can't solve for it?

5. Mar 6, 2006

### Black Orpheus

Alternatively, if I use the fact that the partial of z with respect to x equals -(partial of F with respect to x)/(partial of F with respect to z), don't I still end up with a z (and I need it in terms of x and y so I can plug in my point)?

6. Mar 6, 2006

### codyg1985

Yeah, that trick works as well to find dz/dx.

It will give you something in terms of x and y...after all when taking partials with respect to some variable you will get all of those variables in the result.

Last edited: Mar 6, 2006
7. Mar 6, 2006

### Black Orpheus

If you have time, could you walk me through the dz/dx method? I'm not following this... If I try to take dz/dx with the formula mentioned above, I get -(y + dz/dx + 3z^5)/(dxy/dz + 1 + 15xz^4). Do dz/dx and dxy/dz go to 0? What about the extra zs?

8. Mar 6, 2006

### xman

try just writing in a general sense with letting some function of the variables (x,y) say z=z(x,y) where all the derivative and continuity requirements we need are assumed satisfied. Now from ordinary calc we may write
$$dz = \left( \vec{\nabla} z(x,y) \cdot \hat{x}\right) dx + \left( \vec{\nabla} z(x,y) \cdot \hat{y} \right) dy \Rightarrow dz = \frac{\partial z}{\partial x} dx+ \frac{\partial z}{\partial y} dy$$
would you agree?

Last edited: Mar 7, 2006
9. Mar 6, 2006

### Hurkyl

Staff Emeritus
Why would you need to solve for it? Isn't it enough to just say z can be written as a function of x and y?

I suspect you would profit from reviewing the one-dimensional case of implicit differentiation -- I think you're missing the basic idea behind it, and it might help to work in a simpler setting. (And, presumably, you've learned this once already and it might come back to you. You might even have saved your notes and worked homework problems!)

Last edited: Mar 6, 2006
10. Mar 7, 2006

### codyg1985

If you use the formula:

$$\frac {\partial z}{\partial x} = -\frac {\frac {\partial F}{\partial x}}{\frac {\partial F}{\partial z}}$$

then you will get an answer with only x, y and z in it. The function F has a constant value of zero when you create it by moving everything to one side of the equation. It is still a function of x and y, and z is still the dependant variable.

If you go through with implicit differentiation, you must solve for $$\frac {\partial z}{\partial x}$$, not forgetting to do the product rule and chain rule in places where you have both an x and a z.

In the end you should get the same answer for either method.