Why is this other method of computing a derivative correct? So, normally when we compute the derivative of a certain composite function we simply use a combination of the derivative rules such us chain rule, product rules, etc.. e.g.
$$ \frac{d}{dx} (e^x\sqrt{x^2lnx})$$

Product rule:$\sqrt{x^2lnx}*\frac{d}{dx}e^x +e^x*\frac{d}{dx}\sqrt{x^2lnx}$ 
Chain rule:
$ e^x\sqrt{x^2lnx}+e^x*\frac{1}{2\sqrt{x^2*lnx}}*\frac{d}{dx}x^2*lnx$

Product rule again:$ e^x\sqrt{x^2lnx}+e^x*\frac{1}{2\sqrt{x^2*lnx}}*(2xlnx+x^2*\frac{1}{x})$
Intuitively, the chain rule can be thought as the ratio of the change of each function multiplied, while the product rule can be thought as the change in the area of a square with each function as the side.
But there is another way to compute the derivative as the sum of all the derivatives with respect to the individual isntances of x while holding the other x's constant
$e^x$ ->$e^x*\sqrt{x^2lnx}$

$x^2$->$e^x*\frac{\sqrt{lnx}}{2\sqrt{x^2}}*2x$

$lnx$->$e^x*\frac{\sqrt{x^2}}{2\sqrt{lnx}}*\frac{1}{x}$
And their sum is equal to the derivative of the function.. why is this true intuitively? Does my assumption that it is simply the sum of all changes that each instance of x cause is correct? And does this also work for multivariable funtions or do they have certain limiations?
 A: This is the multivariable chain rule at work, and as long as you're careful keeping everything straight, it will always* give you the correct derivative.
(* assuming niceness, like the function being differentiable in the first place)
Say we have a function of three variables $f(a, b, c)$. We can the differentiate this function with respect to any of the three variables, and we write those partial derivatives like so:
$$
\frac{\partial f}{\partial a}, \frac{\partial f}{\partial b}, \frac{\partial f}{\partial c}
$$
(alternatively $\frac{\partial}{\partial a}f$, etc.) For instance, in this specific case have
$$
f(a, b, c) = e^a\sqrt{b^2\ln c}
$$
which gives
$$
\frac{\partial f}{\partial a}(a, b, c) = e^a\sqrt{b^2\ln c}\\
\frac{\partial f}{\partial b}(a, b, c) = e^a\frac{2b\ln c}{2\sqrt{b^2\ln c}}\\
\frac{\partial f}{\partial c}(a, b, c) = e^a\frac{\sqrt{b^2}/c}{2\sqrt{\ln c}}
$$
Now, if we make each of $a, b, c$ functions of $x$, so that we get a new function
$$
g(x) = f(a(x), b(x), c(x))
$$
then the multivariate chain rule says in general that
$$
\frac{dg}{dx} = \frac{\partial f}{\partial a} \frac{da}{dx} + \frac{\partial f}{\partial b} \frac{db}{dx} + \frac{\partial f}{\partial c} \frac{dc}{dx}
$$
If we specifically set the functions $a(x) = b(x) = c(x) = x$, then we have $\frac{da}{dx} = \frac{db}{dx} = \frac{dc}{dx} = 1$, and we recover your alternate derivation.
You could, if you'd like, instead use something like
$$
f(a, b, c) = e^a\cdot b\cdot\sqrt c\\
a(x) = x, b(x) = \sqrt{x^2}, c(x) = \ln x
$$
and get the same result, as long as you remember that this time the $\frac{d}{dx}$ terms also contribute.
