삼각함수의 미분

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 compute the derivative of $\tan(x)$ as being $\sec^2(x)$, which evaluated at $x=\frac{\pi}{4}$ yields $2$. Since we also compute $\tan(\pi/4)=1$, then the equation of the tangent line is

$$\begin{aligned} y - 1 = 2\cdot (x-\pi/4). \end{aligned}$$

We compute the derivative $y' = \cos(x)-\sin(x)+e^x$, which evaluated at $x=0$ yields $1-0+1 = 2$. Since we also compute $y(0)=0+1+1=2$, the equation of the tangent line is

$$\begin{aligned} y - 2 = 2(x-0) \end{aligned}$$

ie $y=2x+2$.

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 $n$.

First, we note that our function is continuous, because $$\lim_{x \to 0} f(x)=\lim_{x \to 0} \frac{\sin x}{x}=1=f(0)$$ This is a handy thing to check: if the function were discontinuous at $x=0$, then we would automatically know that it was not differentiable there.

Now, on to the derivative. We can use the limit definition:

$$\begin{aligned} f'(0)&=\lim_{h \to 0}\frac{f(0+h)-f(0)}{h}\qquad\text{ if it exists}\\ &=\lim_{h \to 0} \frac{f(h)-1}{h}\\ &=\lim_{h \to 0} \frac{\frac{\sin h}{h}-1}{h} \end{aligned}$$

As $h$ approaches 0, both the numerator and the denominator approach 0. So, to evaluate the limit, we need to do more work. The key insight we can use is a result that was shown in the text while evaluating the derivative of sine. When $h$ is close to 0, $\cos h \leq \frac{\sin h}{h} \leq 1$. We use this to bound our limit, and then apply the squeeze theorem.

$$\begin{aligned} &\cos h&&\leq&&\frac{\sin h}{h}&&\leq&&1\\ \text{So, }\qquad&\cos h-1&&\leq&&\frac{\sin h}{h}-1&&\leq&& 1-1=0 \end{aligned}$$

and $\frac{1}{h}\left(\frac{\sin h}{h}-1\right)$ is between $0$ and $\frac{\cos h-1}{h}$. We can evaluate the limit of $\frac{\cos h-1}{h}$, by noticing its similarity to the definition of the derivative.

$$\begin{aligned} \lim_{h \to 0}\frac{\cos h-1}{h}&=\lim_{h \to 0}\frac{\cos(0+h)-\cos(0)}{h}\\ &=\frac{\mathrm{d}}{\mathrm{d}x}\left.\left\{\cos x \right\}\right|_{x=0}=0 \end{aligned}$$

So, by the Squeeze Theorem, $f'(0) =\displaystyle \lim_{h \to 0}\frac{\frac{\sin h}{h}-1}{h}=0$.

Recall that $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x} =1$. In order to take advantage of this knowledge, we divide the numerator and denominator by $x^5$ (because $5$ is the power of sine in the denominator, and a denominator that goes to zero generally makes a limit harder).

$$\begin{aligned} \lim_{x\rightarrow 0} \dfrac{\sin x^{27}+2x^5 e^{x^{99}}}{\sin^5 x} &=\lim_{x\rightarrow 0} \dfrac{\dfrac{\sin x^{27}}{x^{5}}+2 e^{x^{99}}} {\left(\dfrac{\sin x}{x}\right)^5} \end{aligned}$$

Now the denominator goes to 1, which is nice, but we need to take care of the fraction $\dfrac{\sin x^{27}}{x^5}$ in the numerator. This fraction isn't very familiar, but we know that, as $x$ goes to zero, $x^{27}$ also goes to zero, so that $\dfrac{\sin x^{27}}{x^{27}}$ goes to $1$. Consequently,

$$\begin{aligned} \lim_{x\rightarrow 0} \dfrac{\sin x^{27}+2x^5 e^{x^{99}}}{\sin^5 x} &=\lim_{x\rightarrow 0} \dfrac{x^{22}\dfrac{\sin x^{27}}{x^{27}}+2 e^{x^{99}}} {\left(\dfrac{\sin x}{x}\right)^5} =\dfrac{0\times 1+2\times e^0}{1^5} =2 \end{aligned}$$

(a) January 1, 2012 corresponds to the instant $t = 2$, so we compute $V'(2)$. First, we observe by the sum and constant multiple rules that $V'(t) = 24 \cdot 1.07^t \cdot \ln(1.07) + 6 \cos(t)$. Thus, $V'(2) = 24 \cdot 1.07^2 \cdot \ln(1.07) + 6 \cos(2) \approx -0.63778$. Since $V$ is measured in thousands of dollars and $t$ in years, we know that the portfolio's value is decreasing at a rate of $637.78$ dollars per year on January 1, 2012. (b) Since $V'(t) = 24 \cdot 1.07^t \cdot \ln(1.07) + 6 \cos(t)$, it follows that $V''(t) = 24 \cdot 1.07^t \cdot (\ln(1.07))^2 - 6 \sin(t)$, and thus $V''(2) = 24 \cdot 1.07^2 \cdot \ln(1.07)^2 - 6 \sin(2) \approx -5.33$, with units “thousands of dollars per year per year”. This means that around the time that $t = 2$, the portfolio's value is decreasing at while the value function is concave down. More specifically, we know that $V'$ is decreasing at the moment $t = 2$ and expect the derivative's value to decrease by about $5.33$ thousand dollars per year over the course of the next year. In other words, the portfolio's value is not only decreasing, but we expect it to decrease more rapidly as the year goes on. (c) In the figure below, we see plots of both $V$ and $A$ along with their corresponding derivatives. We see that the function $A$ is a purely exponential function that always increases at an increasing rate. By adding the term $6\sin(t)$ to $A$ to create the function $V$, we add volatility to the value of the portfolio: the oscillating nature of the sine function not only creates variance in the stock's value, but also in its rate of change. We also observe that as time progresses and the value of $24 \cdot (1.07)^t$ increases, the effect of $6 \sin(t)$ lessens.

(a) The slope of the line tangent to the graph of $f$ at $a = \frac{\pi}{4}$ is given by $f'\left(\frac{\pi}{4}\right)$. To find $f'\left(\frac{\pi}{4}\right)$ we first find $f'(x)$, which is $f'(x) = -3\sin(x) - 2\cos(x)$. Thus, $$\begin{aligned}f'\left(\frac{\pi}{4}\right) &= -3\sin\left(\frac{\pi}{4}\right) - 2\cos\left(\frac{\pi}{4}\right) \\ & -3\left(\frac{\sqrt{2}}{2}\right) - 3\left(\frac{\sqrt{2}}{2}\right) = -5\left(\frac{\sqrt{2}}{2}\right)\end{aligned}\,.$$ (b) The linearization $L(x)$ to $f$ at $a = \pi$ is $(x) = f(\pi) + f'(\pi)(x-\pi)$L. We calculated $f'(x)$ in part (a) and so $f'(\pi) = -3\sin(\pi) - 2\cos(\pi) = 2$. We also have $f(\pi) = 3\cos(\pi) - 2\sin(\pi) + 6 = 3$. Hence $L(x) = 3+2(x-\pi)$. (c) A differentiable function $f$ is increasing at $x=a$ if and only if $f'(a) > 0$ and decreasing if and only if $f'(a) < 0$. Now $f'\left(\frac{\pi}{2}\right) = -3\sin\left(\frac{\pi}{2}\right) - 2\cos\left(\frac{\pi}{2}\right) = -3$, so $f$ is decreasing at $a = \frac{\pi}{2}$. (d) If a curve defined by a function $f$ is concave up at $x=a$, then the tangent line to $f$ at $x=a$ lies below the curve and if the graph of $f$ is concave down at $x=a$, then the tangent line to $f$ at $x=a$ lies above the curve. The concavity of a curve is determined by the second derivative and $f''(x) = -3\cos(x)+2\sin(x)$. Since $$f''\left(\frac{3\pi}{2}\right) = -3\cos\left(\frac{3\pi}{2}\right) + 2\sin\left(\frac{3\pi}{2}\right) = -2\,,$$ the graph of $f$ is concave down at $a = \frac{3\pi}{2}$ and the tangent line to $f$ lies above the curve at this point.

(a) By definition, we know that $$f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h}\,.$$ By the sum of two angles identity for the sine function, $\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)$, and thus $$f'(x) = \lim_{h \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}$$ .. Combining the first and last terms and removing a factor of $\sin(x)$, $$f'(x) = \lim_{h \to 0} \frac{ \sin(x)(\cos(h)-1) + \cos(x)\sin(h) - \sin(x) }{h}$$ .. (b) Continuing our work in (a), we divide each part of the numerator by $h$ and see it follows $$f'(x) = \lim_{h \to 0} \left( \frac{\sin(x) (\cos(h)-1) }{h} + \frac{\cos(x)\sin(h)}{h} \right)\,.$$ Next, we note that $\sin(x)$ does not depend on $h$, and likewise $\cos(x)$ is independent of $h$. It thus makes sense to view those constants as being outside the limit being taken, and also that the limit of a sum is the sum of the limits. Hence we write $$f'(x) = \sin(x) \cdot \lim_{h \to 0} \left( \frac{\cos(h)-1}{h} \right) + \cos(x) \cdot \lim_{h \to 0} \left( \frac{\sin(h)}{h} \right)\,.$$ (c) Using small positive and negative values of $h$ to estimate the desired limits, we find that it appears $$\lim_{h \to 0} \left( \frac{\cos(h)-1}{h} \right) = 0$$ and $$\lim_{h \to 0} \left( \frac{\sin(h)}{h} \right) = 1$$ It suffices in each case to consider $h$-values of $h = \pm 0.01, \pm 0.001, \pm 0.0001$. (d) Combining our results in (b) and (c), we have found that $$f'(x) = \sin(x) \cdot \lim_{h \to 0} \left( \frac{\cos(h)-1}{h} \right) + \cos(x) \cdot \lim_{h \to 0} \left( \frac{\sin(h)}{h} \right) = \sin(x) \cdot 0 + \cos(x) \cdot 1 = \cos(x)\,.$$ (e) The argument for showing $\frac{d}{dx}[\cos(x)] = -\sin(x)$ is almost identical. Do note that the sum of two angles identity for $\cos(\alpha + \beta)$ is $\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$.