<s>[INST] Minimum distance between the curves $f(x) =e^x$ and $g(x) =\ln x$ What is the minimum distance between the curves $f(x) =e^x$ and $g(x) = \ln x$?
I didn't understand how to solve the problem. Please help me.
 [/INST] Have a visualization:

The green curve is $f(x) = e^x$, the red curve is $g(x) = \ln(x)$.
$P$ is a point on the graph of $f$, $Q$ is a point on the graph of $g$.
Shown is the distance between $P$ and $Q$.
The obvious symmetry of the graphs results from the fact that $f$ and $g$ are inverse to each other, 
$\DeclareMathOperator{id}{id}f \circ g = g \circ f = \id$ 
which can be constructed as mirroring at the line $y = x$, thus the graph of $\id$.
The distance of the curves is the minimum (perhaps infimum) of all possible distances between a point of the graph of $f$ and a point of the graph of $g$.
\begin{align}
d(f, g) 
&= \min_{x, y} d((x, f(x)), (y, g(y)) \\
&= \min_{x, y} \sqrt{(x - y)^2 + (f(x) - g(y))^2}
\end{align}
With $q = d^2$ we have
$$
q = (x - y)^2 + (e^x - \ln(y))^2
$$
and
$$
\DeclareMathOperator{grad}{grad}
\grad q = (q_x, q_y)
$$
with
\begin{align}
q_x &= 2 (x - y) + 2(e^x - \ln(y)) e^x \\
q_y &= 2 (x - y)(-1) + 2(e^x - \ln(y)) (-(1/y)) 
\end{align}
To find out where the gradient vanishes (from the graph it seems $(0,1)$),
we solve
\begin{align}
0 &= x - y + e^x - \ln(y) \\
0 &= x - y + (e^x - \ln(y))/y
\end{align}
From this we get
$$
0 = (e^x-\ln(y))(1 - 1/y) \iff \\
1 - 1/y = 0 \vee e^x - \ln(y) = 0 \iff \\
y = 1 \vee e^x = \ln(y)
$$
Inserting $y = 1$ we get
$$
0 = x - 1 + e^x \\
1 - x = e^x \\
$$
which means $x = 0$. So $(0,1)$ is a critical point. Going on analytically would mean to look at the second partial derivatives, and check for the positive defniteness at $(0,1)$ to establish the local minimum.
We get
\begin{align}
q_{xx} &= 2 + 2 (e^x)^2 + 2 (e^x - \ln(y)) e^x \\
q_{xy} &= -2 - 2 e^x / y \\
q_{yy} &= 2y + 2 (1/y)^2 + 2(e^x - \ln(y))(1/y^2) \\
\end{align}
so
\begin{align}
q_{xx}(0,1) &= 2 + 2 + 2 = 6 \\
q_{xy}(0,1) &= -4 \\
q_{yy}(0,1) &= 2 + 2 + 2 = 6
\end{align}
and 
$$
\left. H \right|_{(0,1)} = 
\left.
\begin{pmatrix}
q_{xx} & q_{yx} \\
q_{xy} & q_{yy}
\end{pmatrix}
\right|_{(0,1)}
=
\begin{pmatrix}
6 & -4 \\
-4 & 6
\end{pmatrix}
$$
Using linear algebra or a computer algebra system (me lazy) one finds the eigenvalues $2$ and $10$, both positive, so the Hesse matrix is positive definite at $(0,1)$ and it must be a minimum.
</s>