미분법 전체 문제

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

This limit represents the derivative computed at $x=2015$ of the function $f(x)=\cos(x)$. To see this, simply use the definition of the derivative at $a=2015$ with $f(x)=\cos x$:

$$\begin{aligned} \left.\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)\}\right|_{a} &= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}\\ \left.\frac{\mathrm{d}}{\mathrm{d}x}\{\cos x\}\right|_{2015} &= \lim_{x\to2015}\frac{\cos(x)-\cos(2015)}{x-2015} \end{aligned}$$

Since the derivative of $f(x)$ is $-\sin(x)$, its value at $x=2015$ is exactly $-\sin(2015)$.

This limit represents the derivative computed at $x=\pi/3$ of the function $f(x)=\cos x$. To see this, simply use the definition of the derivative at $a=\pi/3$ with $f(x)=\cos x$:

$$\begin{aligned} \left.\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)\}\right|_{a} &= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}\\ \left.\frac{\mathrm{d}}{\mathrm{d}x}\{\cos x\}\right|_{\pi/3} &= \lim_{x\to\pi/3}\frac{\cos(x)-\cos(\pi/3)}{x-\pi/3}\\ &=\lim_{x\to\pi/3} \frac{\cos(x)-1/2}{x-\pi/3} \end{aligned}$$

Since the derivative of $f(x)$ is $-\sin x$, then its value at $x=\pi/3$ is exactly $-\sin(\pi/3)=-\sqrt{3}/2$.

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

$$\begin{aligned} \left.\frac{\mathrm{d}}{\mathrm{d}x}\{f(x)\}\right|_{a} &= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}\\ \left.\frac{\mathrm{d}}{\mathrm{d}x}\{\sin x\}\right|_{\pi} &=\lim_{x\to\pi} \frac{\sin(x)-\sin(\pi)}{x-\pi}\\ &=\lim_{x\to\pi} \frac{\sin(x)}{x-\pi} \end{aligned}$$

Since the derivative of $f(x)$ is $\cos(x)$, then its value at $x=\pi$ is exactly

$\cos(\pi)=-1$.

$$\begin{aligned} \tan \theta &= \dfrac{\sin \theta}{\cos \theta} \\ \text{So, using the quotient rule,} \\ \frac{\mathrm{d}}{\mathrm{d}\theta}\{\tan \theta\}&=\frac{\cos\theta\cos\theta-\sin\theta(-\sin\theta)}{\cos^2\theta} =\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}\\ &=\left(\frac{1}{\cos \theta}\right)^2=\sec^2\theta \end{aligned}$$

In order for the function $f(x)$ to be continuous at $x=0$, the left half formula $ax+b$ and the right half formula $\dfrac{6\cos x}{2+\sin x+\cos x}$ must match up at $x=0$. This forces $$a\times 0+b=\frac{6\cos 0}{2+\sin 0+\cos 0}=\frac{6}{3} \implies \boxed{b=2}$$

In order for the derivative $f'(x)$ to exist at $x=0$, the limit $\displaystyle \lim_{h \rightarrow 0} \dfrac{f(h)-f(0)}{h}$ must exist. In particular, the limits $\displaystyle \lim_{h \rightarrow 0^-} \dfrac{f(h)-f(0)}{h}$ and $\displaystyle \lim_{h \rightarrow 0^+} \dfrac{f(h)-f(0)}{h}$ must exist and be equal to each other.

When $h \to 0^-$, this means $h<0$, so $f(h)=ah+b=ah+2$. So: $$\displaystyle \lim_{h\to 0^-}\dfrac{f(h)-f(0)}{h}=\displaystyle \lim_{h\to 0^-}\dfrac{(ah+2)-2}{h}=\left.\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{ax+2\right\}\right|_{x=0}=a.$$

Similarly, when $h \to 0^+$, then $h>0$, so $f(h)=\dfrac{6\cos h}{1+\sin h + \cos h}$ and $$\begin{aligned} \displaystyle \lim_{h \rightarrow 0^+} \dfrac{f(h)-f(0)}{h} &= \left.\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\dfrac{6\cos x}{2+\sin x + \cos x}\right\}\right|_{x=0} \\&=\left.\frac{-6\sin x(2+\sin x+\cos x)-6\cos x(\cos x-\sin x)}{(2+\sin x+\cos x)^2}\right|_{x=0}.\end{aligned}$$

Since the limits from the left and right must be equal, this forces $$a=\frac{-6\sin 0(2+\sin 0+\cos 0)-6\cos 0(\cos 0-\sin 0)}{(2+\sin 0+\cos 0)^2} =\frac{-6}{(2+1)^2}\implies\boxed{a=-\frac{2}{3}}$$

In order for $f'(x)$ to exist, $f(x)$ has to exist. We already know that $\tan x$ does not exist whenever $x=\frac{\pi}{2}+n\pi$ for any integer $n$. If we look a little deeper, since $\tan x = \frac{\sin x}{\cos x}$, the points where tangent does not exist correspond exactly to the points where cosine is zero.

From its graph, tangent looks like a smooth curve over its domain, so we might guess that everywhere tangent is defined, its derivative is defined. We can check this: $f'(x) = \sec^2 x = \left(\frac{1}{\cos x}\right)^2$. Indeed, wherever $\cos x$ is nonzero, $f'$ exists.

So, $f'(x)$ exists for all values of $x$ except when $x=\frac{\pi}{2}+n\pi$ for some integer $n$.

The function is differentiable whenever $x^2+x-6\ne 0$ since the derivative equals

$$\begin{aligned} \frac{10\cos(x)\cdot (x^2+x-6)-10\sin(x)\cdot (2x+1)}{(x^2+x-6)^2}, \end{aligned}$$

which is well-defined unless $x^2+x-6=0$. We solve $x^2+x-6=(x-2)(x+3)=0,$ and get $x=2$ and $x=-3$. So, the function is differentiable for all real values $x$ except for $x=2$ and for $x=-3$.

The function is differentiable whenever $\sin(x)\ne 0$ since the derivative equals

$$\begin{aligned} \frac{\sin(x)\cdot (2x+6) - \cos(x)\cdot (x^2+6x+5)}{(\sin x)^2}, \end{aligned}$$

which is well-defined unless $\sin x = 0$. This happens when $x$ is an integer multiple of $\pi$. So, the function is differentiable for all real values $x$ except $x=n\pi,$, where $n$ is any integer.

We are asked to solve $f'(x)=0$. That is, $e^x[\sin x + \cos x]=0$. Since $e^x$ is always positive, that means we need to find all points where $\sin x + \cos x =0$. That is, we need to find all values of $x$ where $\sin x = - \cos x$. Looking at the unit circle, we see this happens whenever $x = \frac{3\pi}{4}+n\pi$ for any integer $n$.

At time $t$, the particle is at the point $\big(x(t),y(t)\big)$, with $x(t)=\cos t$ and $y(t)=\sin t$. Over time, the particle traces out a curve; let's call that curve $y=f(x)$. Then $y(t) = f\big(x(t)\big)$, so the slope of the curve at the point $\big(x(t),y(t)\big)$ is $f'\big(x(t)\big)$. You are to determine the values of $t$ for which $f'\big(x(t)\big)=-1$.

By the chain rule

$$\begin{aligned} y'(t) = f'\big(x(t)\big) \cdot x'(t) \end{aligned}$$

Substituting in $x(t)=\cos t$ and $y(t)=\sin t$ gives

$$\begin{aligned} \cos t = f'\big(x(t)\big) \cdot \big(-\sin t\big) \end{aligned}$$

so that

$$\begin{aligned} f'\big(x(t)\big) = -\frac{\cos t}{\sin t} \end{aligned}$$

is $-1$ precisely when $\sin t = \cos t$. This happens whenever $t = \frac{\pi}{4}$.

Remark: the path traced by the particle is a semicircle. You can think about the point on the unit circle with angle t, or you can notice that $x^2 + y^2 = \sin^2t + \cos^2t = 1$.