Which one is the variable? (Derivatives) I'm very confuzzled as to where the variable is here. I don't know where to differentiate. 
Here's the question, differentiate:
$$y = \frac{\sin(\theta)}{2} + \frac{c}{\theta}$$
Do I solve for $f(c)$ or $f(\theta)$? I tried solving it for $f(c)$ treating theta as a constant, but I'm unsure if that's correct.
 A: The correct answer is:

O Questioner, differentiate with respect to what???
Do you want $\frac{dy}{dc}$ or $\frac{dy}{dθ}$? Or perhaps $\frac{dθ}{dc}$?

A: Standard or common notation would have $c$ be the constant, and $\theta$ the variable.
In Calc I, early letters ($a, b, c, d \ldots$) are ususally constants, and late letters ($x, y, z$) are usually variables. $\theta$ is often used for a variable when the variable in question is an angle, just as $t$ is used for time, or $r$ for a radial distance.
A: In strict terms, this is ambiguous. This may be $y(\theta)$ or $y(c)$; in each case the other variable is fixed. However, as noted, the likely intended interpretation is that $\theta$ is the variable. 
A: As x and y are to lengths, heights, quantities, etc., θ and φ are to angles. Moreover, c generally represents some constant.
This is particularly true in physics, where θ often refers to an angle or phase, and c represents the (constant) speed of light.
A: It's a convention to represent constants by the letters c,d,a etc. $\theta$ can be treated as a variable intuitively (especially in Physics). Differentiation is represented in the form $\displaystyle\frac{d}{dx}$ or $\displaystyle\frac{d}{dy}$ or $\displaystyle\frac{d}{dz}$ of something. Here the letter $x,y$ or $z$ in the denominator (denominator is not actually correct as this is not a fraction; but we use it for easing up things) represents the variable.
A: I believe that the intent of the exercise is to differentiate with respect to $\theta$, but you should keep in mind that you may also have to differentiate the function with respect to $c$, or even differentiate with respect to NONE of the variables shown, in which case the value would simply be $0$.
