Let's suppose that we have the function $x^2$. We can write $x^2$ like $$\left(\frac {x^\frac {2}{x}}{1^\frac {1}{x}}\right)^x.$$ Now we may construct a new function: $$\left(\frac {x^\frac {2}{x}}{1^\frac {1}{x}}\right)^{x^3}.$$ Both functions touch each other in the points they meet, that is, their derivatives in those points are the same as well.
Another example. For $2x^2$ the new function is $$\left(\frac{2^{1/x}x^{2/x}}{1^{1/x}}\right)^{2x^3}.$$
In general, the new function of $ax^b$ is $$\left(\frac {a^{1/x}\cdot x^{b/x}}{1^{1/x}}\right)^{ax^{b+1}}. $$
Both touch in their common points.
But why does this rule work? Any algebraic explanation?
There’s a very nice generalization to your question. Your function $$\left(\frac {a^{1/x}\cdot x^{b/x}}{1^{1/x}}\right)^{ax^{b+1}}$$ is actually $(ab^x)^{(ab^x)}$. Therefore let us consider $f(x)^{f(x)}$ for a function $f$ for which $f^f$ does exist.
We will show a fact that was previously unknown at least to me:
We find first that $f(x)^{f(x)}=f(x)$ iff $f(x)=1$, hence $f(x)^{f(x)}=1$ here. In case they meet it must be on the line $y=1$.
The derivative of $f^f$ is calculated as usual to
$$\bigl(f^f\bigr)’=f^f\cdot f’\cdot\bigl(\ln(f)+1\bigr).$$
In each common point we find that $$\bigl(f(x)^{f(x)}\bigr)’=1\cdot f’(x)\bigl(\ln(1)+1\bigr)=f’(x),$$ which was to be shown.
Feel free to draw some pictures, nice ones involve $\sin$ as $f(x)=x\cdot \sin(x)$