다항함수의 미분법

(a) By definition, $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\,.$$ Using the fact that $f(x)=a^x$ along with $a^{x+h} = a^x \cdot a^h$, we have $$\begin{aligned}f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h \to 0} \frac{a^{x+h}-a^x}{h} \\ &= \lim_{h \to 0} \frac{a^{x} \cdot a^{h} - a^x}{h} \\ &= \lim_{h \to 0} \frac{a^{x}(a^{h}-1)}{h}\end{aligned}\,,$$ where the last equality follows by removing the common factor of $a^x$ from each term in the numerator. (b) Since $a^x$ does not depend at all on $h$, we may treat $a^x$ as constant in the noted limit. Thus, $$f'(x) = \lim_{h \to 0} \frac{a^{x}(a^{h}-1)}{h} = a^{x} \cdot \lim_{h \to 0} \frac{a^{h}-1}{h}\,.$$ Note that this tells us that the derivative of $a^x$ is some constant multiple of $a^x$ (and that the constant's values must be $\lim_{h \to 0} \frac{a^{h}-1}{h}$.) (c) Using $h = \pm 0.0001$, it appears that when $a = 2$, $L \approx 0.6931$, while when $a = 3$, $L \approx 1.0986$. (d) By experimenting, we see first that $a \approx 2.7$ (for which $L \approx 0.993$, so $a$ needs to be slightly bigger than $2.7$). Continuing to try values of $a$ that result in $L$ being closer and closer to $1$, we find $a \approx 2.71828$ (for which $L \approx 1.000$) (e) We note that $\ln(2) \approx 0.6931$ and $\ln(3) \approx 1.0986$. Since these values correspond to the values of $$\lim_{h \to 0} \frac{a^{h}-1}{h}$$ when $a = 2$ and $a = 3$, we conjecture that $$\frac{d}{dx}[2^x] = 2^x \cdot \ln(2)$$ and $$\frac{d}{dx}[3^x] = 3^x \cdot \ln(3)$$ (f) Our work suggests that $\frac{d}{dx}[a^x] = a^x \cdot \ln(a)$, and when $a = e \approx 2.71828$, it follows that $$\frac{d}{dx}[e^x] = e^x \cdot \ln(e) = e^x$$ Thus, $e^x$ is a very special function: its derivative is itself!