지수함수와 로그함수의 미분

${x}^{\cot x}\cdot \left[-{\csc }^{2}x\cdot \ln x+\frac{\cot x}{x}\right]$

${x}^{-1/2}{({x}^{2}+3)}^{2/3}{(3x-4)}^{4}\cdot \left[\frac{-1}{2x}+\frac{4x}{3({x}^{2}+3)}+\frac{12}{3x-4}\right]$

a. At the beginning of 1960 there were 5.3 thousand cases of the disease in New York City. At the beginning of 1964 there were approximately 723 cases of the disease in the United States. b. At the beginning of 1960 the number of cases of the disease was decreasing at rate of $-4.611$ thousand per year; at the beginning of 1963, the number of cases of the disease was decreasing at a rate of $-0.2808$ thousand per year.

$p=35741{(1.045)}^{t}$

Using the quotient rule,

$$\begin{aligned} f'(x)=\frac{2xe^x-2e^x}{4x^2}=\frac{e^x(2x-2)}{4x^2}=\frac{(x-1)e^x}{2x^2} \end{aligned}$$

$$\begin{aligned} f'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\{e^{2x}\}=\frac{\mathrm{d}}{\mathrm{d}x}\{(e^x)^2\}=2\frac{\mathrm{d}}{\mathrm{d}x}\{e^x\}e^x=2e^xe^x=2(e^{x})^2=2e^{2x} \end{aligned}$$

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

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 $y=\log x \cdot x^{-3}$, $$y'=\frac{1}{x}x^{-3}+\log x \cdot(-3)x^{-4}=x^{-4}(1-3\log x)$$