극한·미분 심화 전체 문제

Using the chain rule,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\arcsin\left(\frac{x}{3}\right)\right\}&=\frac{1}{\sqrt{1-\left(\frac{x}{3}\right)^2}}\cdot \frac{1}{3}\\ &=\frac{1}{3\sqrt{1-\frac{x^2}{9}}}\\ &=\frac{1}{\sqrt{9-x^2}} \end{aligned}$$

Since the domain of arcsine is $[-1,1]$, and we are plugging in $\dfrac{x}{3}$ to arcsine, the values of $x$ that we can plug in are those that satisfy $-1 \le \dfrac{x}{3} \leq 1$, or $-3\leq x \leq 3$. So the domain of $f$ is $[-3,3]$.

Using the quotient rule,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\left\{\dfrac{\arccos t}{t^2-1}\right\}&= \frac{(t^2-1)\left(\frac{-1}{\sqrt{1-t^2}}\right)-(\arccos t)(2t)}{(t^2-1)^2} \end{aligned}$$

The domain of arccosine is $[-1,1]$, and since $t^2-1$ is in the denominator, the domain of $f$ requires $t^2-1 \neq 0$, that is, $t \neq \pm 1$. So the domain of $f(t)$ is $(-1,1)$.

The domain of $\mathop{\mathrm{arcsec}} x$ is $|x| \geq 1$: that is, we can plug into arcsecant only values with absolute value greater than or equal to one. Since $-x^2-2 \leq -2$, every real value of $x$ gives us an acceptable value to plug into arcsecant. So, the domain of $f(x)$ is all real numbers.

To differentiate, we use the chain rule. Remember $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\mathop{\mathrm{arcsec}} x\right\} = \dfrac{1}{|x|\sqrt{x^2-1}}$.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\mathop{\mathrm{arcsec}}(-x^2-2)\right\}&= \dfrac{1}{|-x^2-2|\sqrt{(-x^2-2)^2-1}}\cdot(-2x)\\ &=\dfrac{-2x}{(x^2+2)\sqrt{x^4+4x^2+3}}. \end{aligned}$$

We use the chain rule, remembering that $a$ is a constant.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{a}\arctan\left(\dfrac{x}{a}\right)\right\}&= \frac{1}{a}\cdot\frac{1}{1+\left(\frac{x}{a}\right)^2}\cdot \frac{1}{a}\\ &=\frac{1}{a^2+x^2} \end{aligned}$$

The domain of arctangent is all real numbers, so the domain of $f(x)$ is also all real numbers.

We differentiate using the product and chain rules.

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{ x\arcsin x + \sqrt{1-x^2}\right\}&= \arcsin x + \frac{x}{\sqrt{1-x^2}}+ \frac{-2x}{2\sqrt{1-x^2}}\\ &=\arcsin x \end{aligned}$$

The domain of $\arcsin x$ is $[-1,1]$, and the domain of $\sqrt{1-x^2}$ is all values of $x$ so that $1-x^2 \geq 0$, so $x$ in $[-1,1]$. Therefore, the domain of $f(x)$ is $[-1,1]$.

We differentiate using the chain rule:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\{\arctan(x^2)\}&=\frac{2x}{1+x^4} \end{aligned}$$

This is zero exactly when $x=0$.

Using formulas you should memorize from this section, $$\frac{\mathrm{d}}{\mathrm{d}x}\{\arcsin x + \arccos x\}=\frac{1}{\sqrt{1-x^2}}+\frac{-1}{\sqrt{1-x^2}}=0$$

Remark: the only functions with derivative equal to zero everywhere are constant functions, so $\arcsin x + \arccos x$ should be a constant. Since $\sin \theta = \cos \left(\frac{\pi}{2}-\theta\right)$, we can set

$$\begin{aligned} \sin\theta&=x & \cos\left(\frac{\pi}{2}-\theta\right)&=x\\ \text{where $x$ and $\theta$ are the same in both expressions, and $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$. Then} \\ \arcsin x &=\theta & \arccos x &= \frac{\pi}{2}-\theta \\ \text{We note here that arcsine is the inverse of the sine function \textit{ restricted to} $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. So, since we restricted $\theta$ to this domain, $\sin \theta=x$ really does imply $\arcsin x = \theta$. (For an example of why this matters, note $\sin(2\pi)=0$, but $\arcsin (0)=0 \neq 2\pi$.) Similarly, arccosine is the inverse of the cosine function restricted to $[0,\pi]$. Since $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, then $0 \leq (\frac{\pi}{2}-\theta) \leq \pi$, so $\cos\left(\frac{\pi}{2}- \theta\right) =x$ really does imply $\arccos x=\frac{\pi}{2}-\theta$.} \\ \end{aligned}$$

So, $$\arcsin x+\arccos x =\theta+\frac{\pi}{2}-\theta =\frac{\pi}{2}$$ which means the derivative we were calculating was actually just $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\dfrac{\pi}{2}\right\}=0$.

Using the chain rule, $$y'=\frac{-{\frac{1}{ x^2}}}{\sqrt{1-\left({\frac{1}{ x}}\right)^2}}=\frac{-1}{x^2\sqrt{1-\frac{1}{x^2}}}.$$

Using the chain rule, $$y'=\frac{-{\frac{1}{ x^2}}}{1+\left({\frac{1}{ x}}\right)^2}=\frac{-1}{x^2+1}.$$

Using the product rule:

$$\begin{aligned} \displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{(1+x^2)\arctan x\right\} &=2x\arctan x+(1+x^2)\frac{1}{1+x^2}\\ &=2x\arctan x+1\end{aligned}$$

The line $y=2x+9$ has slope $2$, so we must find all values of $x$ between $-1$ and $1$ ($\arcsin x$ is only defined for these values of $x$) for which $\frac{\mathrm{d}}{\mathrm{d}x}\{\arcsin x\}=2$. Evaluating the derivative:

$$\begin{aligned} y&=\arcsin x\\ 2=y'&=\frac{1}{\sqrt{1-x^2}}\\ 4&=\frac{1}{1-x^2}\\ \frac{1}{4}&=1-x^2\\ x^2&=\frac{3}{4}\\ x&=\pm\frac{\sqrt{3}}{2}\\ (x,y)&=\pm\big(\frac{\sqrt{3}}{2},\frac{\pi}{3}\big) \end{aligned}$$

We differentiate using the chain rule:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\{\arctan(\csc x)\}&=\frac{1}{1+\csc^2x}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{\csc x\}\\ &=\frac{-\csc x \cot x}{1+\csc^2x}\\ &=\frac{-\frac{1}{\sin x}\cdot \frac{\cos x}{\sin x}}{1+\left(\frac{1}{\sin x}\right)^2}\\ &=\frac{-\cos x}{\sin^2x+1} \end{aligned}$$

So if $f'(x)=0$, then $\cos x=0$, and this happens when $x=\dfrac{(2n+1)\pi}{2}$ for any integer $n$. We should check that these points are in the domain of $f$. Arctangent is defined for all real numbers, so we only need to check the domain of cosecant; when $x=\dfrac{(2n+1)\pi}{2}$, then $\sin x=\pm1 \neq 0$, so $\csc x = \dfrac{1}{\sin x}$ exists.

We approximate the function $f(x)=\sqrt{x}$ about $x=4$, since 4 is a perfect square and it is close to 5.

$$\begin{aligned} f(4)&=\sqrt{4}=2\\ f'(x)&=\frac{1}{2\sqrt{x}} \qquad \Rightarrow \qquad f'(4)=\frac{1}{2\sqrt{4}}=\frac{1}{4}\\ f(x) &\approx f(4)+f'(4)(x-4)=2+\frac{1}{4}(x-4)\\ f(5)&\approx 2+\frac{1}{4}(5-4)=\frac{9}{4} \end{aligned}$$

We estimate $\sqrt{5}\approx\dfrac{9}{4}$.

Remark: $\left(\dfrac{9}{4}\right)^2=\dfrac{81}{16}$, which is pretty close to $\dfrac{80}{16}=5$. Our approximation seems pretty good.

We approximate the function $f(x)=\sqrt[5]{x}$. We need to centre the approximation about some value $x=a$ such that we know $f(a)$ and $f'(a)$, and $a$ is not too far from $30$.

$$\begin{aligned} f(x)&=\sqrt[5]{x}=x^{\frac{1}{5}}\\ f'(x)&=\frac{1}{5}x^{-\frac{4}{5}}=\frac{1}{5\sqrt[5]{x}^4} \\ \text{$a$ needs to be a number whose fifth root we know. Since $\sqrt[5]{32}=2$, and 32 is reasonably close to 30, $a=32$ is a great choice.} \\ f(32)&=\sqrt[5]{32}=2\\ f'(32)&=\frac{1}{5\cdot 2^4}=\frac{1}{80} \\ \text{The linear approximation of $f(x)$ about $x=32$ is} \\ f(x) &\approx 2+\frac{1}{80}(x-32) \\ \text{When $x=30$:} \\ f(30) &\approx 2+\frac{1}{80}(30-32)=2-\frac{1}{40}=\frac{79}{40} \end{aligned}$$

We estimate $\sqrt[5]{30}\approx \dfrac{79}{40}$.

Remark: $\dfrac{79}{40} = 1.975$, while $\sqrt[5]{30} \approx 1.97435$. This is a decent estimation.

First, let's find the first and second derivatives of $f$ when $x=5$.

$$\begin{aligned} f'(x)&=\frac{1}{1+x^2} & f'(5)&=\frac{1}{26}\\ f''(x)&=\frac{-2x}{\left(1+x^2\right)^2} & f''(5)&=\frac{-10}{26^2} \end{aligned}$$

$$\begin{aligned} \\ \text{The linear approximation for $\Delta y$ when $\Delta x = \dfrac{1}{10}$ is} \\ \Delta y &\approx f'(5)\Delta x = \frac{1}{26}\cdot \frac{1}{10}=\frac{1}{260} \approx 0.003846 \\ \text{The quadratic approximation for $\Delta y$ when $\Delta x = \dfrac{1}{10}$ is} \\ \Delta y &\approx f'(5)\Delta x + \frac{1}{2}f''(5)\left(\Delta x\right)^2 = \frac{1}{26}\cdot\frac{1}{10}+\frac{1}{2}\cdot\frac{-10}{26^2}\cdot\frac{1}{10^2}=\frac{51}{13520}\approx 0.003772 \end{aligned}$$

Remark: $\Delta y = \arctan(5.1)-\arctan(5)\approx 0.003774$.

If we plug in $x=1$ to the numerator and the denominator, we find they are both zero. So, we have an indeterminate form appropriate for L'Hopital's Rule.

$$\begin{aligned} \lim\limits_{x\rightarrow 1}\underbrace{\frac{x^3-e^{x-1}}{\sin(\pi x)}}_{\genfrac{}{}{0in}{}{\mathrm{num} \to 0}{\mathrm{den} \to 0}} =\lim\limits_{x\rightarrow 1}\frac{3x^2-e^{x-1}}{\pi\cos(\pi x)} =-\frac{2}{\pi}\end{aligned}$$

Be careful– this is not an indeterminate form!

As $x\rightarrow 0+$, the numerator $\log x\rightarrow-\infty$. That is, the numerator is becoming an increasingly huge, negative number. As $x\rightarrow 0+$, the denominator $x\rightarrow 0+$, which only serves to make the total fraction even larger, and still negative. So, $\lim\limits_{x\rightarrow 0^+}\dfrac{\log x}{x} =-\infty$.

Remark: if we had tried to use l'Hopital's Rule here, we would have come up with the wrong answer. If we differentiate the numerator and the denominator, the fraction becomes $\dfrac{\frac{1}{x}}{1}=\frac{1}{x}$, and $\displaystyle \lim_{x \to 0^+}\frac{1}{x}=\infty$. The reason we cannot apply l'Hopital's Rule is that we do not have an indeterminate form, like both numerator and denominator going to infinity, or both numerator and denominator going to zero.

We rearrange the expression to a more natural form:

$$\begin{aligned} \lim_{x\rightarrow\infty}(\log x)^2e^{-x} &=\lim_{x\rightarrow\infty}\underbrace{\dfrac{(\log x)^2}{e^{x}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}} \\ \text{Both the numerator and denominator go to infinity as $x$ goes to infinity. So, we can apply l'H\^opital's Rule. In fact, we end up applying it twice.} \\ &=\lim_{x\rightarrow\infty}\underbrace{\dfrac{2\log x}{xe^{x}}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}} \\ &=\lim_{x\rightarrow\infty}\dfrac{2/x}{xe^{x}+e^x} \\ \text{The numerator gets smaller and smaller while the denominator gets larger and larger, so:} \\ &={0} \end{aligned}$$