미분법 전체 문제

Possible points of inflection: $x=-\pi/4,3\pi/4$; concave down on $(-\pi/4,3\pi/4)$ concave up on $(-\pi,-\pi/4)$ and $(3\pi/4,\pi)$

Possible points of inflection: $x=-2\pm\sqrt{2}$; concave down on $(-2-\sqrt{2},-2+\sqrt{2})$ concave up on $(-\infty,-2-\sqrt{2})$ and $(-2+\sqrt{2},\infty)$

Possible points of inflection: $x=1/e^{3/2}$; concave down on $(0,1/e^{3/2})$ concave up on $(1/e^{3/2},\infty)$

Possible points of inflection: $x=\pm 1/\sqrt{2}$; concave down on $(- 1/\sqrt{2}, 1/\sqrt{2})$ concave up on $(-\infty,- 1/\sqrt{2})$ and $( 1/\sqrt{2},\infty)$

(a) $5/(2\pi)\approx 0.796$cm/s (b) $1/(40\pi)\approx 0.00796$ cm/s (c) $1/(4000\pi)\approx 0.0000796$ cm/s

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 $a=b=\dfrac{e}{2}$.

You should memorize the derivatives of sine, cosine, and tangent.

$f'(x)=\cos x - \sin x + \sec^2 x$

$f'(x)=\cos x - \sin x$, so $f'(x)=0$ precisely when $\sin x = \cos x$. This happens at $\pi/4$, but it also happens at $5\pi/4$. By looking at the unit circle, it is clear that $\sin x = \cos x$ whenever $x = \frac{\pi}{4}+\pi n$ for some integer $n$.

• Solution 1: $f(x)=\sin^2x+\cos^2x=1$, so $f'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\{1\}=0$.
• Solution 2: Using the formula for the derivative of a squared function,

$$\begin{aligned} f'(x)=2\sin x \cos x + 2\cos x(- \sin x)=2\sin x \cos x - 2 \sin x \cos x =0. \end{aligned}$$

It is true that $2\sin x \cos x = \sin (2x)$, but we don't know the derivative of $\sin(2x)$. So, we use the product rule: $$f'(x)=2\cos x \cos x+2\sin x (-\sin x)=2(\cos^2 x - \sin ^2 x).$$

We use the quotient rule.

$$\begin{aligned} f'(x)&=\dfrac{(\cos x + \tan x)(2\cos x + 3 \sec^2 x)-(2\sin x+3\tan x)(-\sin x + \sec^2 x)}{(\cos x + \tan x)^2}\\ &=\frac{2\cos^2x+3\cos x \sec^2 x + 2 \cos x \tan x + 3 \tan x \sec^2 x}{(\cos x + \tan x)^2}\\ &~~~~ +\frac{2\sin^2x-2\sin x \sec^2x+3\sin x \tan x -3\tan x \sec^2 x}{(\cos x + \tan x)^2}\\ &=\frac{2+3 \sec x + 2 \sin x -2\tan x \sec x+3\sin x \tan x }{(\cos x + \tan x)^2} \end{aligned}$$

We use the quotient rule.

$$\begin{aligned} f'(x) &= \frac{e^x(5\sec x \tan x)-(5\sec x + 1) e^x}{(e^{x})^2}\\ &=\frac{5\sec x \tan x - 5 \sec x - 1}{e^x} \end{aligned}$$

We use the product rule:

$$\begin{aligned} f'(x)&=(e^x+\cot x)(30x^5+\csc x \cot x)+(e^x-\csc^2x)(5x^6-\csc x) \end{aligned}$$

We don't know how to differentiate this function as it is written, but an identity helps us. Since $\sin\left(\frac{\pi}{2}-\theta \right)=\cos \theta$, we see $f'(\theta)=\frac{\mathrm{d}}{\mathrm{d}\theta}\{\cos \theta\}=-\sin(\theta)$.