$6 \cdot 63$ When the graph $G$ of $y=\lg x$ is rotated counterclockwise by $90^{\circ}$ about the origin, a new graph $G^{\prime}$ is obtained. Which of the following equations represents $G^{\prime}$?
(A) $y=\lg \left(\frac{x+90}{9}\right)$.
(B) $y=\log _{x} 10$.
(C) $y=\frac{1}{x+1}$.
(D) $y=10^{-x}$.
(E) $y=10^{x}$.
(42nd American High School Mathematics Examination, 1991)
[Solution 1] Clearly, the graphs of $y=f(x)$ and $x=f(y)$ are symmetric about the line $y=x$.
Also, the graphs of the function $-x=f(y)$ and $x=f(y)$ are symmetric about the $y$-axis.
Therefore, the graph of the function $-x=f(y)$ can be seen as the result of rotating the graph of the function $y=f(x)$ counterclockwise by $90^{\circ}$ around the origin $O$ in the $x O y$ plane.
And $-x=f(y)$ is commonly written as $y=f^{-1}(-x)$.
$\therefore$ The required equation is $y=10^{-x}$.
Thus, the answer is $(D)$.
[Solution 2] By the properties of complex numbers in the Gaussian plane:
The complex number representation of the point $(x, \lg x)$ is $x+i \lg x$, and after rotating this point counterclockwise by $90^{\circ}$ around the origin, the new point is
$$
\begin{array}{l}
Z=(x+i \lg x) i=-\lg x+i x, \quad \text { or }(-\lg x, x) . \\
\text { Let } \quad X=-\lg x, \quad Y=x \text {, then } \\
Y=x=10^{-x} .
\end{array}
$$

Thus, the answer is $(D)$.