Contradiction when differentiating? Consider the function $F = x+y$. Let $x = t$ and $y= \cos t$. By directly differentiating,
$$\frac{\partial F}{\partial x} = 1$$ and $$\frac{\partial F}{\partial y} = 1$$
Using the chain rule however,
$$\frac{\partial F}{\partial x} = \frac{\partial{F}}{\partial {x}}\frac{dx}{dx} + \frac{\partial{F}}{\partial {y}}\frac{dy}{dx}$$ $$= 1-\sin x$$
How is this possible?
 A: Your result from directly differentiating is incorrect.  $y$ is a function of $t$, which is equal to $x$.  So $y$ is a function of $x$.  Therefore, the derivative of $F=x+y$  with respect ot $x$ is the result from the chain rule.
In this case, you could write $F$ as $F=x + \cos(x)$, which may make things more clear.
A: $F(x,y)=x+y$ defines $F$ to be a function with domain $\mathbb{R}^2$ and range $\mathbb{R}$ that takes $(a,b)$ to $a+b$. With $F$ defined this way, $\frac{\partial F}{\partial x}$ is the derivative of $F$ with respect to the first coordinate (the one labeled $x$ in the definition), and it’s the function on $\mathbb{R}^2$ that has constant value 1.
But as soon as you “let $x=t$ and $y=\cos t$,” the meaning of $\frac{\partial F}{\partial x}$ becomes unclear. Is it a derivative of a function on $\mathbb{R}^2$ with respect to the first coordinate, or is it the derivative of the function of one variable $x$ that takes $x$ to $x+\cos x$?
It would be nice if standard mathematical notation were clearer about this, but it’s not. 
Here’s a nice essay on this topic.
