극한·미분 심화 전체 문제

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.

Since $g(y)=f^{-1}(y)$,

$$\begin{aligned} f(g(y))&=f\left(f^{-1}(y)\right)=y \\ \text{Now, we can differentiate with respect to $y$ using the chain rule.} \\ \frac{\mathrm{d}}{\mathrm{d}y}\left\{f(g(y))\right\}&=\frac{\mathrm{d}}{\mathrm{d}y}\{y\}\\ f'(g(y))\cdot g'(y)&=1\\ g'(y)&=\frac{1}{f'(g(y))}=\frac{1}{1-\sin g(y)} \end{aligned}$$

Write $g(y)=f^{-1}(y)$. Then $g(f(x))=x$, so differentiating both sides (using the chain rule), we see

$$\begin{aligned} g'(f(x))\cdot f'(x)=1 \\ \text{What we want is $g'(\pi-1)$, so we need to figure out which value of $x$ gives $f(x)=\pi-1$. A little trial and error leads us to $x=\frac{\pi}{2}$.} \\ g'(\pi-1)\cdot f'\left(\frac{\pi}{2}\right)&=1 \\ \text{Since $f'(x)=2-\cos(x)$, $f'\left(\frac{\pi}{2}\right)=2-0=2$:} \\ g'(\pi-1)\cdot 2&=1\\ g'(\pi-1)=\frac{1}{2} \end{aligned}$$

Write $g(y)=f^{-1}(y)$. Then $g(f(x))=x$, so differentiating both sides (using the chain rule), we see

$$\begin{aligned} g'(f(x))f'(x)&=1 \\ \text{What we want is $g'(e+1)$, so we need to figure out which value of $x$ gives $f(x)=e+1$. A little trial and error leads us to $x=1$.} \\ g'(f(1))f'(1)&=1\\ g'(e+1)\cdot f'(1)&=1\\ g'(e+1) &= \frac{1}{f'(1)} \\ \text{It remains only to note that $f'(x)=e^x+1$, so $f'(1)=e+1$} \\ g'(e+1)&=\frac{1}{e+1} \end{aligned}$$

We use logarithmic differentiation, our standard method of differentiating an expression of the form $(\text{function})^{\text{function}}$.

$$\begin{aligned} f(x)&=[\sin x +2]^{\mathop{\mathrm{arcsec}} x}\\ \log(f(x))&=\mathop{\mathrm{arcsec}} x \cdot \log[\sin x +2]\\ \frac{f'(x)}{f(x)}&=\frac{1}{|x|\sqrt{x^2-1}}\log[\sin x +2]+\mathop{\mathrm{arcsec}} x \cdot \frac{\cos x}{\sin x +2}\\ f'(x)&=[\sin x +2]^{\mathop{\mathrm{arcsec}} x}\left(\frac{\log[\sin x +2]}{|x|\sqrt{x^2-1}}+ \frac{\mathop{\mathrm{arcsec}} x \cdot\cos x}{\sin x +2}\right) \end{aligned}$$

The domain of $\mathop{\mathrm{arcsec}} x$ is $|x| \geq 1$. For any $x$, $\sin x +2$ is positive, and a positive number can be raised to any power. (Recall negative numbers cannot be raised to any power–for example, $(-1)^{1/2}=\sqrt{-1}$ is not a real number.) So, the domain of $f(x)$ is $|x| \geq 1$.

The function $\dfrac{1}{\sqrt{x^2-1}}$ exists only for those values of $x$ with $x^2-1>0$: that is, the domain of $\dfrac{1}{\sqrt{x^2-1}}$ is $|x|>1$. However, the domain of arcsine is $|x| \leq 1$. So, there is not one single value of $x$ where $\arcsin x$ and $\dfrac{1}{\sqrt{x^2-1}}$ are both defined.

If the derivative of $\arcsin(x)$ were given by $\dfrac{1}{\sqrt{x^2-1}}$, then the derivative of $\arcsin(x)$ would not exist anywhere, so we would probably just write “derivative does not exist," instead of making up a function with a mismatched domain. Also, the function $f(x)=\arcsin(x)$ is a smooth curve–its derivative exists at every point strictly inside its domain. (Remember not all curves are like this: for instance, $g(x)=|x|$ does not have a derivative at $x=0$, but $x=0$ is strictly inside its domain.) So, it's a pretty good bet that the derivative of arcsine is not $\dfrac{1}{\sqrt{x^2-1}}$.

This limit represents the derivative computed at $x=1$ of the function $f(x)=\arctan x$. To see this, simply use the definition of the derivative at $a=1$:

$$\begin{aligned} \left.\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)\}\right|_{a} &= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}\\ \left.\frac{\mathrm{d}}{\mathrm{d}x}\{\arctan x\}\right|_{1} &=\lim_{x\to1} \frac{\arctan x-\arctan 1}{x-1}\\ &=\lim_{x\to1} \frac{\arctan x-\frac{\pi}{4}}{x-1}\\ &=\displaystyle \lim_{x\to 1}\left( (x-1)^{-1}\left(\arctan x - \frac{\pi}{4}\right)\right). \end{aligned}$$

Since the derivative of $f(x)$ is $\dfrac{1}{1+x^2}$, its value at $x=1$ is exactly $\dfrac{1}{2}$.

First, let's interpret the given information: when the input of our function is $2x+1$ for some $x$, then its output is $\dfrac{5x-9}{3x+7}$, for that same $x$. We're asked to evaluate $f^{-1}(7)$, which is the number $y$ with the property that $f(y)=7$. If the output of our function is 7, that means

$$\begin{aligned} 7&=\frac{5x-9}{3x+7} \\ \text{and so} \\ 7(3x+7)&=5x-9\\ x&=-\frac{29}{8} \\ \text{So, when $x=-\dfrac{29}{8}$, our equation $f(2x+1)=\dfrac{5x-9}{3x+7}$ becomes:} \\ f\left(2\cdot\frac{-29}{8}+1\right)&=\dfrac{5\cdot\frac{-29}{8}-9}{3\cdot\frac{-29}{8}+7} \\ \text{Or, equivalently:} \\ f\left(-\frac{25}{4}\right)&=7 \end{aligned}$$

Therefore, $f^{-1}(7)=-\dfrac{25}{4}$.

If $f^{-1}(y)=0$, that means $f(0)=y$. So, we want to find out what we plug into $f^{-1}$ to get 0. Since we only know $f^{-1}$ in terms of a variable $x$, let's figure out what $x$ gives us an output of 0:

$$\begin{aligned} \frac{2x+3}{x+1}&=0\\ 2x+3&=0\\ x&=-\frac{3}{2} \\ \text{Now, the equation $f^{-1}(4x-1)=\dfrac{2x+3}{x+1}$ with $x=\dfrac{-3}{2}$ tells us:} \\ f^{-1}\left(4\cdot\frac{-3}{2}-1\right)&=\frac{2\cdot\frac{-3}{2}+3}{\frac{-3}{2}+1} \\ \text{Or, equivalently:} \\ f^{-1}(-7)&=0 \end{aligned}$$

Therefore, $f(0)=-7$.

• Solution 1: We begin by differentiating implicitly. Following the usual convention, we use $y'$ to mean $y'(x)$.

$$\begin{aligned} \arcsin(x+2y)&=x^2+y^2 \qquad\text{Using the chain rule:}\\ \frac{1+2y'}{\sqrt{1-(x+2y)^2}}&=2x+2yy'\\ \frac{1}{\sqrt{1-(x+2y)^2}}+\frac{2y'}{\sqrt{1-(x+2y)^2}}&=2x+2yy'\\ \frac{2y'}{\sqrt{1-(x+2y)^2}}-2yy'&=2x-\frac{1}{\sqrt{1-(x+2y)^2}}\\ y'\left(\frac{2}{\sqrt{1-(x+2y)^2}}-2y\right)&=2x-\frac{1}{\sqrt{1-(x+2y)^2}}\\ y'&=\frac{2x-\frac{1}{\sqrt{1-(x+2y)^2}}}{\frac{2}{\sqrt{1-(x+2y)^2}}-2y} \left(\frac{\sqrt{1-(x+2y)^2}}{\sqrt{1-(x+2y)^2}}\right)\\ y'&=\frac{2x\sqrt{1-(x+2y)^2}-1}{2-2y\sqrt{1-(x+2y)^2}} \end{aligned}$$

• Solution 2: We begin by taking the sine of both sides of the equation.

$$\begin{aligned} \arcsin(x+2y)&=x^2+y^2\\ x+2y&=\sin(x^2+y^2) \\ \text{Now, we differentiate implicitly.} \\ 1+2y'&=\cos(x^2+y^2)\cdot(2x+2yy')\\ 1+2y'&=2x\cos(x^2+y^2)+2yy'\cos(x^2+y^2)\\ 2y'-2yy'\cos(x^2+y^2)&=2x\cos(x^2+y^2)-1\\ y'\left(2-2y\cos(x^2+y^2)\right)&=2x\cos(x^2+y^2)-1\\ y'&=\frac{2x\cos(x^2+y^2)-1}{2-2y\cos(x^2+y^2)} \end{aligned}$$

• We used two different methods, and got two answers that look pretty different. However, the answers ought to be equivalent. To see this, we remember that for all values of $x$ and $y$ that we care about (those pairs $(x,y)$ in the domain of our curve), the equality $$\arcsin(x+2y)=x^2+y^2$$ holds. Drawing a triangle:

where the adjacent side (in red) come from the Pythagorean Theorem. Then, $\cos(x^2+y^2)=\sqrt{1-(x+2y)^2}$, so using our second solution:

$$\begin{aligned} y'&=\frac{2x\cos(x^2+y^2)-1}{2-2y\cos(x^2+y^2)}\\ &=\frac{2x\sqrt{1-(x+2y)^2}-1}{2-2y\sqrt{1-(x+2y)^2}} \end{aligned}$$

which is exactly the answer from our first solution.

Since $10.1$ is pretty close to $10$, we estimate $10.1^3 \approx 10^3=1000$.

Remark: these kinds of approximations are very useful when you are doing computations. It's easy to make a mistake in your work, and having in mind that $10.1^3$ should be about a thousand is a good way to check that whatever answer you have makes sense.

The linear approximation is $L(x)=3x-9$. Since we're approximating at $x=5$, $f(5)=L(5)$, and $f'(5)=L'(5)$. However, there is no guarantee that $f(x)$ and $L(x)$ have the same value when $x \neq 5$. So:

(a) $f(5)=L(5)=6$

(b) $f'(5)=L'(5)=3$

(c) there is not enough information to find $f(0)$.