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$\frac{d}{dx}\psi(x,y)=\frac{∂\psi}{∂x}+\frac{∂\psi}{∂y}\frac{dy}{dx}$ .

My interpretation is that the left-hand side is the partial derivative of x.

The right hand is the partial derivative of x, plus the partial derivative of y times the derivative of y with respect to x.

I am absolutely confused. Which is the partial derivative, and what does y have to do with x, aren't they BOTH independent variables??

Doesn't y being expressed as a function of x defeat the entire purpose of a multivariable function??

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    What is $P\,$? Guess you meant: $$\frac{d}{dx}\psi\big(x,y(x)\big)=\frac{∂\psi}{∂x}+\frac{∂\psi}{∂y}\frac{dy}{dx}$$ The LHS is a [total derivative](https://en.wikipedia.org/wiki/Total_derivative) in $x\,$, not a partial one. – 2017-02-21
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    @dxiv i wrote it in terms of P and found out that there was a psi symbol lol – 2017-02-21

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In my opinion this is much more clear if we don't suppress inputs to functions. Here it is in different notation:

Suppose $f(x) =\Psi(x,y(x))$ for all real numbers $x$. Then $f'(x) = D_1\Psi(x,y(x)) + D_2 \Psi(x,y(x)) y'(x)$ for all real $x$.

Here $D_1 \Psi$ is the partial derivative of $\Psi$ with respect to its first input, and $D_2 \Psi$ is the partial derivative of $\Psi$ with respect to its second input. I am assuming $\Psi$ and $y$ are differentiable functions.

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    This makes SO much sense. Thanks! :) -- but must we always assume y is a differentiable function of x?? – 2017-02-21
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    Sure. What else would $y$ be? Certainly $y$ is a function, and if $y$ is not differentiable then we are not guaranteed that $f$ is differentiable. – 2017-02-21
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    So, I'm kind of visualizing it like the product rule, from the way you showed it. take the partial derivative of x and add it to the partial derivative of y, and then apply chain rule, so tack on a y' – 2017-02-21
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    In my mind, we are simply using the multivariable chain rule to compute the derivative of $f$. I was not thinking about the product rule at all when I wrote this. – 2017-02-21