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I am a first year undergraduate student planning to major in physics and I am taking a first year course in basic mathematics. We are currently studying partial differentiation.

While going through the course materials, there is something which I was not totally comfortable with.

We have a function $x = r\cos\theta$ where $x$, $r$, and $\theta$ are all variables with $x$ as the dependent variable. What we want to find is $\partial r/\partial x$ and $\partial\theta /\partial x$.

What the lecturer has done is to simply partially differentiate the entire thing with respect to $x$ itself:

$$x = r \cos\theta$$

$$\frac{\partial x}{\partial x}= \frac{\partial}{\partial x}(r\cos\theta)$$

This is where my problem arises. When we take a partial derivative don't we treat everything other than our differentiating variable as constant? Why do we end up with what's below?

$$1 = r\frac{\partial}{\partial x}(\cos\theta) + \cos\theta \frac{\partial r}{\partial x}$$

$$1 = -r\sin\theta\frac{\partial\theta}{\partial x} + \cos\theta\frac{\partial r}{\partial x}$$

Thank you in advance.

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  • $\begingroup$ If "we treat everything other than our differentiating variable as constant", then you seem to think that implicitly partially differentiating $x = r \cos \theta$ with respect to $x$ should give $1 = 0$, which is self-evidently false. When you write "$x = r \cos \theta$" and then treat $x$ as an independent variable, you force $r$ and $\theta$ to be dependent on $x$. ($r$ and $\theta$ will also be dependent on $y$.) So is your "we treat everything other than our differentiating variable as constant" exactly what you mean, or did you mean something close to that? $\endgroup$ – Eric Towers Nov 3 '19 at 3:13
  • $\begingroup$ Yeah I meant something close to that. Sorry, this is kind of new to me. I knew that in some way r and θ are dependent on x. Their dependence on y is something I did not think about too much. Thanks $\endgroup$ – Siddharth Yajaman Nov 3 '19 at 4:14
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The lecturer appears to be treating $r$ and $\theta$ as functions of $x$ and $y$. So think $r=r(x,y)$ and $\theta=\theta(x,y)$. In that case, he is considering $$x=r(x,y)\cos\theta(x,y)$$ and applying $\tfrac{\partial}{\partial x}$ to both sides.

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  • $\begingroup$ Thanks a lot. Makes a lot more sense now. $\endgroup$ – Siddharth Yajaman Nov 3 '19 at 4:14

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