지수함수와 로그함수의 미분

Since $1^x=1$ for any $x$, we see that $(b)$ is just the constant function $y=1$, so D matches to $(b)$.

Since $2^{-x}=\frac{1}{2^x}=\left(\frac{1}{2}\right)^x$, functions $(a)$ and $(d)$ are the same. This is the only function out of the lot that grows as $x\to-\infty$ and shrinks as $x \to \infty$, so A matches to $(a)$ and $(d)$.

This leaves B and C to match to $(c)$ and $(e)$. Since $3>2$, when $x>0$, $3^x>2^x$. So, $(e)$ matches to the function that grows more quickly to the right of the $x$-axis: B matches to $(e)$, and C matches to $(c)$.

First, let's consider the behaviour of exponential functions $a^x$ based on whether $a$ is greater or less than 1. As we know, $\displaystyle \lim_{x\to\infty}a^x=\left\{\begin{array}{ll} \infty & a>1\\0&a<1 \end{array}\right.$ and $\displaystyle \lim_{x\to-\infty}a^x=\left\{\begin{array}{ll} 0 & a>1\\\infty&a<1 \end{array}\right.$. Our function has $\displaystyle \lim_{x \to \infty} f(x)=\infty$ and $\displaystyle \lim_{x \to -\infty} f(x)=0$, so we conclude $a>1$: thus $(d)$ and also $(b)$ hold. (We could have also seen that $(b)$ holds because $a^x$ is defined for all real numbers.)

It remains to decide whether $a$ is greater or less than $e$. (If $a$ were equal to $e$, then $f'(x)$ would be the same as $f(x)$.) We saw in the text that $\frac{\mathrm{d}}{\mathrm{d}x}\{a^x\}=C(a)a^x$ for the function $C(a)=\displaystyle \lim_{h \to 0} \dfrac{a^h-1}{h}$. We know that $C(e)=1$. (Actually, we chose $e$ to be the number that has this property.) From our graph, we see that $f'(x)<f(x)$, so $C(a)<1=C(e)$. In other words, $\displaystyle \lim_{h \to 0} \dfrac{a^h-1}{h}<\displaystyle \lim_{h \to 0} \dfrac{e^h-1}{h}$; so, $a<e$. Thus $(e)$ holds.