Suppose we're tying to differentiate the function $f(x)=x^x$. Now the textbook method would be to notice that $f(x)=e^{x \log{x}}$ and use the chain rule to find $$f'(x)=(1+\log{x})\ e^{x \log{x}}=(1+\log{x})\ x^x.$$

But suppose that I didn't make this observation and instead tried to apply the following differentiation rules:

$$\frac{d}{dx}x^c=cx^{c-1} \qquad (1)\\ \frac{d}{dx}c^x = \log{c}\ \cdot c^x \quad (2)$$

which are valid for any constant $c$. Obviously neither rules are applicable to the form $x^x$ because in this case neither the base nor the exponent are constant. But if I *pretend* that the exponent is constant and apply rule $(1)$, I would get $f'(x)\stackrel{?}{=}x\cdot x^{x-1}=x^x.$ Likewise, if I pretend that the base is constant and apply rule $(2)$ I obtain $f'(x)\stackrel{?}{=}\log{x}\cdot x^x$.

It isn't hard to see that neither of the derivatives are correct. But here's where the magic happens: if we sum the two “derivatives” we end up with
$$x^x+ \log{x}\cdot x^x=(1+\log{x})\ x^x$$
which *is* the correct expression for $f'(x)$.

This same trick yields correct results in other contexts as well. In fact, in some cases it turns out to be a more efficient way of taking derivatives. For example, consider $$g(x)=x^2 = \color{blue} x\cdot \color{red} x.$$ If we pretend the blue $\color{blue} x$ is a constant we would get $g'(x)\stackrel{?}{=}\color{blue}x\cdot 1=x$. Now if we pretend the red $\color{red}x$ is constant we get $g'(x)\stackrel{?}{=}1\cdot \color{red} x=x$. Summing both expressions we end up with $2x$ which is of course a correct expression for the derivative.

These observations have led me to the following conjecture:

> 
> Let $f(x,y)$ be a differentiable function mapping $\mathbb{R}^2$ to $\mathbb{R}.$ Let $f'\_1 (x,y)=\frac{\partial}{\partial x} f(x,y)$ and $f'\_2 (x,y)=\frac{\partial}{\partial y} f(x,y)$. Then for any $t$ we have: $$\frac{d}{dt}f(t,t)=f'\_1 (t,t) + f'\_2 (t,t).$$
> 
> 
> 

(I apologise for the somewhat awkward notation which I could not seem to get around without causing undue ambiguity.)

This formulation also seems to lend itself to following generalisation:

> 
> Let $f:\mathbb{R}^N \to \mathbb{R}$ be a function differentiable in each of its variables $x\_1,x\_2,\ldots,x\_N$. For $n=1,2,\ldots,N$ define $f'\_n(x\_1,x\_2,\ldots,x\_N)=\frac{\partial}{\partial x\_n}f(x\_1,x\_2,\ldots,x\_N)$. Let $t$ be any real number and define the $N$-tuple $T=(t,t,\ldots,t)$. Then one has: $$\frac{d}{dt} f(T)=\sum\_n f'\_n(T).$$
> 
> 
> 

Thus my question is:

* Is this true?
* How can it be proven? (Specifically in the case $N=2$ but also in the general case.)