지수함수와 로그함수의 미분

$2x{e}^{x}+{x}^{2}{e}^{x}$

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.

The power rule tells us that $\frac{\mathrm{d}}{\mathrm{d}x}\{x^n\}=nx^{n-1}$. In this equation, the variable is the base, and the exponent is a constant. In the function $e^x$, it's reversed: the variable is the exponent, and the base it a constant. So, the power rule does not apply.

We are given that one speaker produces 3dB. So if $P$ is the power of one speaker,

$$\begin{aligned}3=V(P)&=10\log_{10}\left(\frac{P}{S}\right). \\ \text{So, for ten speakers:} \\ V(10P)&=10\log_{10}\left(\frac{10P}{S}\right)=10\log_{10}\left(\frac{P}{S}\right)+10\log_{10}\left(10\right)\\ &=3+10(1)=13 \mathrm{dB} \\ \text{and for one hundred speakers:} \\ V(100P)&=10\log_{10}\left(\frac{100P}{S}\right)=10\log_{10}\left(\frac{P}{S}\right)+10\log_{10}\left(100\right)\\ &=3+10(2)=23 \mathrm{dB} \end{aligned}$$

The investment doubles when it hits $2000. So, we find the value of $t$ that gives $A(t)=2000$:

$$\begin{aligned} 2000&=A(t)\\ 2000&=1000e^{t/20}\\ 2&=e^{t/20}\\ \log 2 &=\frac{t}{20}\\ 20\log2&=t \end{aligned}$$

From our logarithm rules, we know that when $y$ is positive, $\log (y^2)=2\log y$. However, the expression $\cos x$ does not always take on positive values, so (a) is not correct. (For instance, when $x=\pi$, $\log(\cos^2 x)=\log(\cos^2\pi)=\log\left((-1)^2\right) = \log(1)=0$, while $2\log (\cos \pi)=2\log(-1)$, which does not exist.)

Because $\cos^2 x$ is never negative, we notice that $\cos^2 x = |\cos x|^2$. When $\cos x$ is nonzero, $|\cos x|$ is positive, so our logarithm rules tell us $\log\left(|\cos x|^2\right) =2\log|\cos x |$. When $\cos x$ is exactly zero, then both $\log(\cos^2x)$ and $2\log|\cos x|$ do not exist. So, $\log(\cos^2x) = 2\log|\cos x|$.