미분법 전체 문제

$\lim_{\theta \to \pi }\frac{\sin \theta }{\tan \theta }=\frac{\sin \pi }{\tan \pi }=\frac{0}{0};$ then, $\lim_{\theta \to \pi }\frac{\sin \theta }{\tan \theta }=\lim_{\theta \to \pi }\frac{\sin \theta }{\frac{\sin \theta }{\cos \theta }}=\lim_{\theta \to \pi }\cos \theta =-1$

$\frac{dy}{dx}=2x-\sec x\tan x$

$\frac{4}{{(5-2x)}^{3}}$

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

$\frac{-3}{5}$

1

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.

The graph $f(x)=\sin x$ has horizontal tangent lines precisely at those points where $\cos x=0$. This must be true, since $\frac{\mathrm{d}}{\mathrm{d}x}\{\sin x\}=\cos x$: where the derivative of sine is zero, cosine itself is zero.

The graph $f(x)=\sin x$ has maximum slope at those points where $\cos x$ has a maximum. This makes sense, because $f'(x)=\cos x$: the maximum values of the slope of sine correspond to the maximum values of cosine.

$$\frac{\mathrm{d}A}{\mathrm{d}E}=\frac{\mathrm{d}A}{\mathrm{d}B}\cdot\frac{\mathrm{d}B}{\mathrm{d}C}\cdot\frac{\mathrm{d}C}{\mathrm{d}D}\cdot\frac{\mathrm{d}D}{\mathrm{d}E}<0$$ since we multiply three positive quantities and one negative.

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

We use the power rule (a) and the chain rule (b): the power rule tells us to “bring down the 2", and the chain rule tells us to multiply by $y'$.

There is no need for the quotient rule here, as there are no quotients. Exponential functions have the form $(\text{constant})^{\text{function}}$, but our function has the form $(\text{function})^{\text{constant}}$, so we did not use (d).

At $(0,4)$ and $(0,-4)$, the curve looks to be horizontal, if you zoom in: a tangent line here would have derivative zero. At the origin, the curve looks like its tangent line is vertical, so $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ does not exist.

(a) No. A function must pass the vertical line test: one input cannot result in two (or more) outputs. Since one value of $x$ sometimes corresponds to two values of $y$ (for example, when $x=\pi/4$, $y$ is $\pm 1/\sqrt{2}$), there is no function $f(x)$ so that $y=f(x)$ captures every point on the circle.

Remark: $y=\pm\sqrt{1-x^2}$ does capture every point on the unit circle. However, since one input $x$ sometimes results in two outputs $y$, this expression is not a function.

(b) No, for the same reasons as (a). If $f'(x)$ is a function, then it can give at most one slope corresponding to one value of $x$. Since one value of $x$ can correspond to two points on the circle with different slopes, $f'(x)$ cannot give the slope of every point on the circle. For example, fix any $0<a<1$. There are two points on the circle with $x$-coordinate equal to $a$. At the upper one, the slope is strictly negative. At the lower one, the slope is strictly positive.

(c) We differentiate:

$$\begin{aligned} 2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \\ \text{and solve for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{x}{y} \end{aligned}$$

But there is a $y$ in the right-hand side of this equation, and it's not clear how to get it out. Our answer in (b) tells us that, actually, we can't get it out, if we want the right-hand side to be a function of $x$. The derivative cannot be expressed as a function of $x$, because one value of $x$ corresponds to multiple points on the circle.