Interesting Differentiation Technique @HansEngler Left the following response to this question regarding "bad math" that works,


  Here's another classical freshman calculus example: 
  
  Find $\frac{d}{dx}x^x$. 
  
  Alice says "this is like  $\frac{d}{dx}x^n = nx^{n-1}$, so the answer is $x x^{x-1} = x^x$."
  Bob says "no, this is like  $\frac{d}{dx}a^x = \log a \cdot a^x$, so the answer is $\log x \cdot x^x$." 
  Charlie says "if you're not sure, just add the two terms, so you'll get partial credit".
  
  The answer  $\frac{d}{dx}x^x = (1 + \log x)x^x $ turns out to be correct.


In this comment, @joriki asserts that this is not "bad math" but rather a legitimate technique,


  You get the derivative of any expression with respect to $x$ as the sums of all the derivatives with respect to the individual instances of $x$ while holding other instances constant.


I had never previously seen such a technique so naturally I tested it on a few examples, including $\frac{d}{dx} \left( x^{ \sin x}\right)$, etc. and it provided the correct result. The following three questions arose,

$ \ \ $  1. What is the proof of its is validity? 
$ \ \ $  2. Are there any examples where this technique outshines standard methods?
Look at $y = f(u(x),v(x))$:
$$\frac{dy}{dx} = \frac{\partial f}{\partial u}\cdot\frac{du}{dx}+\frac{\partial f}{\partial v}\cdot\frac{dv}{dx}$$
Now, note that $x^{\sin x}$ can be written as:
$$y=x^{\sin x} = f(x,\sin x), \ f(u,v) = u^v$$
So that:
$$\frac{dy}{dx} = \frac{\partial f}{\partial u}\cdot\frac{du}{dx}+\frac{\partial f}{\partial v}\cdot\frac{dv}{dx}=
vu^{v-1} \cdot\frac{du}{dx} + \log u \cdot u^v \cdot\frac{dv}{dx}$$
In fact, this holds for any number of terms (since so does the chain rule), where each time we only look at the derivative by $x$ of only one of the terms, multiply those by the "internal" derivative, and sum them all up.
I beg to differ with the suggested differential to the x to the power of x. The result would be best expressed using ln x or the log to base e of x. Ie 

d/dx(x ^x) = x ^x (ln x + 1)

When using log(believed to be in base 10), the result will be expressed as

d/dx(x ^x) = x^x ln10(log x + log e)

Problems of this form could all be solved by taking the log or ln of both sides and implicitly differentiating with respect  to x