Partial Implicit Differentiation

[mattmns]

Just wondering if I did this right:

Here is the question: find $$\frac{\partial z}{\partial x} of \frac{x^2}{9} - \frac{y^2}{4} + \frac{z^2}{2} = 1$$

Now I put the $$\frac{\partial z}{\partial x}$$ on both sides then got.

$$\frac{2x}{9} - 0 + z \frac{\partial z}{\partial x} = 0$$

So


$$\frac{\partial z}{\partial x} = -\frac{2x}{9z}$$

Now I know this is the right answer I am just curious if I did it right, it has been a while.

Now my reasoning:

$$\frac{\partial z}{\partial x} \frac{x^2}{9}$$ is just the derivative with respect to x, so it will be $$\frac{2x}{9}$$

$$\frac{\partial z}{\partial x} \frac{y^2}{4}$$ has no z or x, so it is constand and therefore 0.

$$\frac{\partial z}{\partial x} \frac{z^2}{2}$$ has a z, so it is the derivative of itself, but times $$\frac{\partial z}{\partial x}$$

Is all of that correct, or did I do something wrong. The book I have only shows one example

Thanks!

 

[dextercioby]

It looks okay.You could have done it using the theorem of implicit functions,as well.

Daniel.

 

[mattmns]

Thanks dex! I have no idea what the theorem of implicit functions is, but I probably could not use it anyway.

 

[mattmns]

One more question for now

I need to find

$$\frac{\partial z}{\partial x} sin(xz)$$

How exactly does this work?

Will it be: cos(xz) [times] what? $$\frac{\partial_}{\partial x} xz$$

Which will then be cos(xz) [times] x ? This one seems to be wrong. Any ideas?