지수함수와 로그함수의 미분

$$\begin{aligned} e^{a+x}&=e^ae^x \\ \text{Since $e^a$ is just a constant,} \\ \frac{\mathrm{d}}{\mathrm{d}x}\{e^{a}e^{x}\}&=e^a\frac{\mathrm{d}}{\mathrm{d}x}\{e^x\}=e^ae^x=e^{a+x} \end{aligned}$$

So, $f'(x)=f(x)=e^{a+x}$.

$$\begin{aligned}e^{-x}&=\frac{1}{e^x} \\ \text{Using the rule for differentiating the reciprocal:} \\ \frac{\mathrm{d}}{\mathrm{d}x}\{e^{-x}\}&=\frac{-e^x}{(e^x)^2}=\frac{-1}{e^x}=-e^{-x} \end{aligned}$$

The question asks when $s'(t)$ is negative. So, we start by differentiating. Using the product rule:

$$\begin{aligned} s'(t)&=e^t(t^2+2t)\\ &=e^t \cdot t(t+2) \end{aligned}$$

$e^t$ is always positive, so $s'(t)$ is negative when $t$ and $2+t$ have opposite signs. This occurs when $-2<t<0$.

We simplify the functions to get a better idea of what's going on.

($a$): $y=e^{3\log x}+1=\left(e^{\log x}\right)^3+1=x^3+1$. This is not a line.

($b$): $2y+5=e^{3+\log x}=e^3e^{\log x}=e^3x$. Since $e^3$ is a constant, $2y+5=e^3x$ is a line.

($c$): There isn't a fancy simplification here–this isn't a line. If that isn't a satisfactory answer, we can check: a line is a function with a constant slope. For our function, $y'=\frac{\mathrm{d}}{\mathrm{d}x}\{e^{2x} +4\}=\frac{\mathrm{d}}{\mathrm{d}x}\{e^{2x}\}=\frac{\mathrm{d}}{\mathrm{d}x}\left\{(e^x)^2\right\}=2e^xe^x=2e^{2x}$. Since the derivative isn't constant, the function isn't a line.

($d$): $y=e^{\log x}3^e+\log 2=3^e\,x+\log 2$. Since $3^e$ and $\log 2$ are constants, this is a line.

When we say a function is differentiable without specifying a range, we mean that it is differentiable over its domain. The function $f(x)$ is differentiable when $x \neq 1$ for any values of $a$ and $b$; it is up to us to figure out which constants make it differentiable when $x=1$.

In order to be differentiable, a function must be continuous. The definition of continuity tells us that, for $f$ to be continuous at $x=1$, we need $\displaystyle \lim_{x \to 1}f(x)=f(1)$. From the definition of $f$, we see $f(1)=a+b=\displaystyle \lim_{x \to 1^-}f(x)$, so we need $\displaystyle \lim_{x\to 1^+}f(x)=a+b$. Since $\displaystyle \lim_{x \to 1^+}f(x)=e^1=e$, we specifically need $$e=a+b.$$

Now, let's consider differentiability of $f$ at $x=1$. We need the following limit to exist:

$$\begin{aligned} \lim_{h \to 0} \frac{f(1+h)-f(1)}{h}& \\ \text{In particular, we need the one-sided limits to exist and be equal:} \\ \lim_{h \to 0^-}\frac{f(1+h)-f(1)}{h}&= \lim_{h \to 0^+}\frac{f(1+h)-f(1)}{h} \\ \text{If $h<0$, then $1+h<1$, so $f(1+h)=a(1+h)^2+b$. If $h>0$, then $1+h>1$, so $f(1+h)=e^{1+h}$. With this in mind, we begin to evaluate the one-sided limits:} \\ \lim_{h \to 0^-}\frac{f(1+h)-f(1)}{h}&= \lim_{h \to 0^-}\frac{[a(1+h)^2+b]-[a+b]}{h}\\ &=\lim_{h \to 0^-}\frac{ah^2+2ah}{h}=2a\\ \lim_{h \to 0^+}\frac{f(1+h)-f(1)}{h}&= \lim_{h \to 0^+}\frac{e^{1+h}-(a+b)}{h} \\ \text{Since we take $a+b$ to be equal to $e$ (to ensure continuity):} \\ &= \lim_{h \to 0^+}\frac{e^{1+h}-e^1}{h}\\ &=\left.\frac{\mathrm{d}}{\mathrm{d}x}\{e^x\}\right|_{x=1}=e^1=e \end{aligned}$$

So, we also need $$2a= e$$

Therefore, the values of $a$ and $b$ that make $f$ differentiable are $a=b=\dfrac{e}{2}$.

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, $\log(\cos^2x) = 2\log|\cos x|$.

• Solution 1: Using the chain rule, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(10x)\right\}=\dfrac{1}{10x}\cdot 10=\frac{1}{x}$.
• Solution 2: Simplifying, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(10x)\right\}=\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(10)+\log x\right\} = 0+\frac{1}{x}=\frac{1}{x}$.

• Solution 1: Using the chain rule, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(x^2)\right\}=\frac{1}{x^2}\cdot 2x = \frac{2}{x}$.
• Solution 2: Simplifying, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(x^2)\right\}= \displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{2\log(x)\right\}= \frac{2}{x}$.

Don't be fooled by a common mistake: $\log(x^2+x)$ is not the same as $\log(x^2)+\log x$.

We differentiate using the chain rule: $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\log(x^2+x)\right\}=\dfrac{1}{x^2+x}\cdot(2x+1) = \dfrac{2x+1}{x^2+x}$.

We know the derivative of the natural logarithm (base $e$), so we use the base-change formula:

$$\begin{aligned} f(x)=\log_{10}x&=\frac{\log x}{\log 10} \\ \text{Since $\log 10$ is a constant:} \\ f'(x)&=\frac{1}{x\log 10}. \end{aligned}$$

• Solution 1: Using the quotient rule, $$y'=\frac{x^3\frac{1}{x}-(\log x)\cdot 3x^2}{x^6}= \frac{x^2-3x^2\log x}{x^6}=\frac{1-3\log x}{x^4}.$$
• Solution 2: Using the product rule with $y=\log x \cdot x^{-3}$, $$y'=\frac{1}{x}x^{-3}+\log x \cdot(-3)x^{-4}=x^{-4}(1-3\log x)$$

Using the chain rule,

$$\begin{aligned} \displaystyle \frac{\mathrm{d}}{\mathrm{d}\theta} \log(\sec \theta)&=\frac{1}{\sec \theta}\cdot (\sec \theta \cdot \tan \theta)\\ &=\tan\theta \end{aligned}$$

Remark: the domain of the function $\log(\sec \theta)$ is those values of $\theta$ for which $\sec\theta$ is positive: so, the intervals $\left(\left(2n-\frac{1}{2}\right)\pi,\left(2n+\frac{1}{2}\right)\pi\right)$ where $n$ is any integer. Certainly the tangent function has a larger domain than this, but outside the domain of $\log(\sec \theta)$, $\tan \theta$ is not the derivative of $\log(\sec \theta)$.

Let's start in with the chain rule.

$$\begin{aligned} f'(x) &= e^{ \cos\left(\log x\right)} \cdot \frac{\mathrm{d}}{\mathrm{d}x} \left\{ \cos\left( \log x \right)\right\} \\ \text{We'll need the chain rule again:} \\ &= e^{\cos\left(\log x\right)} (-\sin( \log x)) \cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ \log x \}\\ &= e^{\cos\left(\log x\right)} (-\sin(\log x)) \cdot \frac{1}{x}\\ &=\frac{-e^{\cos(\log x)}\sin(\log x)}{x} \end{aligned}$$

Remark: Although we have a logarithm in the exponent, we can't cancel. The expression $e^{\cos (\log x)}$ is not the same as the expression $x^{\cos x}$, or $\cos x$.

$$\begin{aligned} y&=\log(x^2+\sqrt{x^4+1}) \\ \text{So, we'll need the chain rule:} \\ y'&=\frac{\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^2+\sqrt{x^4+1}\right\}}{ x^2+\sqrt{x^4+1}}\\ &=\frac{2x+\frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{x^4+1}\right\}}{x^2+\sqrt{x^4+1}} \\ \text{We need the chain rule again:} \\ &=\frac{2x+\frac{\frac{\mathrm{d}}{\mathrm{d}x}\left\{ x^4+1\right\}}{2\sqrt{ x^4+1}}}{x^2+\sqrt{x^4+1}}\\ &=\frac{2x+\frac{4x^3}{2\sqrt{x^4+1}}}{x^2+\sqrt{x^4+1}}. \end{aligned}$$

This requires us to apply the chain rule twice.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \left\{ \sqrt{-\log(\cos x)} \right\} &= \frac{1}{2\sqrt{ -\log(\cos x)}} \cdot \frac{\mathrm{d}}{\mathrm{d}x} \left\{ -\log\left(\cos x\right) \right\}\\ &= -\frac{1}{2\sqrt{-\log(\cos x)}} \cdot \frac{1}{ \cos x} \frac{\mathrm{d}}{\mathrm{d}x} \left\{ \cos x \right\}\\ &= -\frac{1}{2\sqrt{-\log(\cos x)}} \cdot \frac{1}{\cos x} \cdot \left(-\sin x\right)\\ &=\frac{\tan x}{2\sqrt{-\log(\cos x)}} \end{aligned}$$

Remark: it looks strange to see a negative sign in the argument of a square root. Since the cosine function always gives values that are at most 1, $\log(\cos x)$ is always negative or zero over its domain. So, $\sqrt{\log(\cos x)}$ is only defined for the points where $\cos x=1$ (and so $\log(\cos x) = 0$–this isn't a very interesting function! In contrast, $-\log(\cos x)$ is always positive or zero over its domain – and therefore we can always take its square root.