Why Are r and x Derivative Relationships Not Contradictory?

[dtl42]

Homework Statement

Why are the following not contradictory?
r=\sqrt{x^2+y^2}
\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}=\frac{x}{r}=cos{\theta}

and

r=\frac{x}{cos{\theta}}
\frac{\partial r}{\partial x}=\frac{1}{cos{\theta}}

Homework Equations

The Attempt at a Solution

I understand that we got these results from differentiating two separate relations, but I don't see why they're different, and seemingly contradictory.

 

[gabbagabbahey]

Hint: what is the definition of \theta?...If \theta has any explicit dependence on x, then is \frac{\partial}{\partial x} \left(\frac{x}{cos(\theta)} \right) Really just \frac{1}{cos(\theta)}?Don't you have to use the product rule?

 

[HallsofIvy]

They are contradictory because the second is simply wrong. Yes,
r= x/cos(\theta) but the derivative is wrong: \theta is not a constant, it depends on x itself. A correct calculation would be
\frac{\partial r}{\partial x}= \frac{cos(\theta)+ xsin(\theta)\frac{\partial \theta}{\partial x}}{cos^2(\theta}[/itex]