Find the derivative (Implicit)

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hoch449
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Homework Statement



Find [tex]\frac{\partial\theta}{\partial y}[/tex]

[tex]z=rcos\theta[/tex]
[tex]x=rsin\theta\cos\phi[/tex]
[tex]y=rsin\theta\sin\phi[/tex]
[tex]r^2=x^2 + y^2 + z^2[/tex]


The Attempt at a Solution



We know [tex]cos\theta=\frac{z}{r}=\frac{z}{\sqrt{x^2 + y^2 + z^2}}[/tex]

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!
 
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An earlier part to this question was to find [tex]\frac{\partial r}{\partial y}[/tex] and I solved it correctly.

Here is how I did it.

[tex]r^2= x^2 + y^2 + z^2[/tex]

[tex]\frac{d}{dy}r^2= \frac{d}{dy}y^2[/tex]

[tex]\frac{d}{dr}r^2\frac{dr}{dy}=2y[/tex]

[tex]2r\frac{dr}{dy}=2y[/tex]

so therefore [tex]\frac{\partial r}{\partial y}= \frac{y}{r}[/tex]

I am just having some difficulty with the next part of the question.
 
hoch449 said:

The Attempt at a Solution



We know [tex]cos\theta=\frac{z}{r}=\frac{z}{\sqrt{x^2 + y^2 + z^2}}[/tex]

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

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

[tex]\frac{\partial\theta}{\partial y}:[/tex]

[tex]\frac{d}{dy}cos\theta=\frac{d}{dy}(\frac{z}{r})[/tex]

[tex]\frac{d}{d\theta}(cos\theta)\frac{d\theta}{dy}=\frac{d}{dr}r^-1\frac{dr}{dy}[/tex]

[tex]-sin\theta\frac{d\theta}{dy}=-r^-2\frac{dr}{dy}[/tex]

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