Conjecture about Cal 1 derivatives? 
Conjecture:
Let $F\left(\vec{x}\right) : \Bbb{R}^n \to \Bbb{R}$
Define $g(t) = F(t, t, \dots, t)$
Then $$g^{\prime} (t) = \left(\sum_{i=1}^n \ { \partial F \over \partial
 x_i}\right)\Big( \ \{ \ \vec{x} = \langle t,t,\dots,t\rangle \Big)$$


This theorem, if true, could greatly simplify finding the derivatives of most cal 1 functions, and functions in general. 
My first observation of this pattern was in finding the derivative of $x^x$ Informally, this function has two parts - a part which behaves like $x^a$, and a part which behaves like $a^x$. Treating $a$ as a constant and deriving these two functions with respect to $x$ yields $ax^{a-1}$ and $a^x \ln x$. Plugging $x$ back in for $a$ in these yields $x^x$ and $x^x \ln x$. Separately, these derivatives are incorrect, but their sum, $x^x + x^x \ln x$, is the correct derivative.
This pattern held for several examples that I verified by hand, including $x^2 = x \cdot x $, $2x = x + x$, and even the function $\log_x (x)$, which is identical to $1$, yields the correct derivative $0$ when derived this method.
And yet I don't know how to prove the conjecture in general. However, contained within this method are several other basic derivative rules. For example, the product rule and quotient rule can be proven with this method.
Product rule proof: let $g(t) = a(t)b(t)$. Define $F(x,y) = a(x)b(y)$
Then ${\partial F \over \partial x} = a^{\prime}(x) \cdot b(y)\\ {\partial F \over \partial y} = b^{\prime}(y) \cdot a(x)$
So $g^{\prime}(t)= {\partial F \over \partial x} (t,t) + {\partial F \over \partial y} (t,t) = a^{\prime}(t)  \cdot b(t) + b^{\prime}(t) \cdot a(t)$
Similar proof for the quotient rule.
This conjecture also must be true for all $F$ which are linear combinations of smaller, univariate functions,  $F = \sum_i \ a_i \ f_i (x_i)$, because then this method demonstrates the linearity of the derivative operation.
This conjecture can also prove the power rule, ${d \over dx} x^n = nx^{n-1}$
Proof: Define $F(\vec{x}) = \prod_{i=1}^n \ x_i$, and thus $g(t) := F(t,t,\dots,t) = t^n$
Then $${\partial F \over \partial x_i} =\prod_{1 \ \le \ r \ \le \ n, \ \ r \ \ne \ i} \ \ x_r$$  Thus $${\partial F \over \partial x_i}(t,\dots,t) = t^{n-1} $$
So $$g^{\prime}(t) = \sum_i { \partial F \over \partial x_i} (t,\dots,t)  = \sum_{i=1}^n t^{n-1} = n \ t^{n-1}$$
I am very confident that this theorem could be true in general, but I can't think of a way to prove it in general. My professor was unconvinced when I showed him, and neither of us could think of any prior theorem this conjecture might be calling to. Is this conjecture true in general, or does there exist a counterexample function?
 A: This is a special case of the multivariable chain rule. 
Nice discovery! 
If you want a puzzle, you could try to find the derivative of
\begin{equation}
g(t) = F(h_1(t),\ldots,h_n(t)).
\end{equation}
Here $F:\mathbb R^n \to \mathbb R$ and $h_i:\mathbb R \to \mathbb R$ for $i = 1,\ldots,n$.  We can assume that $F$ has continuous partial derivatives and that each function $h_i$ is differentiable. 
(If you don't want the puzzle but just want a full explanation, then you can read about the multivariable chain rule.)
A: @littleO my solution to your puzzle was too large for a comment.
I define a new path function $\vec{h} (t) : \Bbb{R} \to \Bbb{R}^n$,   $\ \ \vec{h}(t) = \langle h_1(t), h_2(t), \dots,h_i(t) \rangle$
Then $g(t)$ can be expressed as the composed function $F\left(\vec{h}\right)$ and thus, using chain rule, 
$g^\prime (t) =\  $D$F(\vec{h}) \ \cdot \ $D$\vec{h}$, where D is the partial derivative matrix from cal 3.
This has matrix representation 
$$\left[ {\partial F \over \partial x_1} \cdots {\partial F \over \partial x_1} \right] \cdot \left[ \begin{array}{c} \ h_1^\prime(t) \\ \vdots \\ h_i^\prime (t) \end{array} \right]$$
Simple matrix multiplication yields $$g^\prime(t) = \sum_{n=1}^i {\partial F \over \partial x_n}(\vec{h}) \cdot h_n^\prime(t)$$
Which proves my conjecture, as $\vec{h} = \langle t,\dots,t \rangle$, and thus $h_n^\prime(t) = 1$
Thank you!!
