# Find the derivative (Implicit)

1. Jun 1, 2009

### hoch449

1. The problem statement, all variables and given/known data

Find $$\frac{\partial\theta}{\partial y}$$

$$z=rcos\theta$$
$$x=rsin\theta\cos\phi$$
$$y=rsin\theta\sin\phi$$
$$r^2=x^2 + y^2 + z^2$$

3. The attempt at a solution

We know $$cos\theta=\frac{z}{r}=\frac{z}{\sqrt{x^2 + y^2 + z^2}}$$

So implicit differentiation says to differentiate both sides with respect to y and this is where I begin to run into trouble.

Please be very specific when you try to explain how this is done lol.

Thanks!

2. Jun 1, 2009

### rock.freak667

Do you know how to do implicit differentiation? Do you know how find the partial derivative of a function with respect to a variable?

3. Jun 1, 2009

### hoch449

An earlier part to this question was to find $$\frac{\partial r}{\partial y}$$ and I solved it correctly.

Here is how I did it.

$$r^2= x^2 + y^2 + z^2$$

$$\frac{d}{dy}r^2= \frac{d}{dy}y^2$$

$$\frac{d}{dr}r^2\frac{dr}{dy}=2y$$

$$2r\frac{dr}{dy}=2y$$

so therefore $$\frac{\partial r}{\partial y}= \frac{y}{r}$$

I am just having some difficulty with the next part of the question.

4. Jun 1, 2009

### slider142

In more specific terms, differentiate both sides of the equation with respect to y keeping all variables other than y and those that are explicit functions of y constant.
Where specifically did you run into trouble?

5. Jun 1, 2009

### hoch449

The Right Hand Side of the equation gives me the difficulty. I am sure I am making an elementary mistake.

$$\frac{\partial\theta}{\partial y}:$$

$$\frac{d}{dy}cos\theta=\frac{d}{dy}(\frac{z}{r})$$

$$\frac{d}{d\theta}(cos\theta)\frac{d\theta}{dy}=\frac{d}{dr}r^-1\frac{dr}{dy}$$

$$-sin\theta\frac{d\theta}{dy}=-r^-2\frac{dr}{dy}$$

I have a feeling that I have already made a mistake with the right side...