삼각함수의 미분

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

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 $n$.

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) 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) We use the quotient, product, sum, and constant multiple rules to obtain $$\begin{aligned}p'(z) &= \frac{d}{dz}\left(\frac{z\tan(z)}{z^2\sec(z) + 1} + 3 e^z + 1\right) \\ &= \frac{d}{dz}\left(\frac{z\tan(z)}{z^2\sec(z) + 1}\right) + \frac{d}{dz} \left(3 e^z \right) + \frac{d}{dz}(1) \\ &= \frac{\left(z^2\sec(z) +1 \right)\frac{d}{dz}\left(z\tan(z)\right) - z\tan(z) \frac{d}{dz}\left(z^2\sec(z) + 1\right)}{\left(z^2\sec(z) + 1\right)^2} +3e^z \\ &= \frac{\left(z^2\sec(z) +1 \right)\left(z\sec^2(z)+\tan(z)\right) - z\tan(z) \left(z^2\sec(z)\tan(z)+2z\sec(z)\right)}{\left(z^2\sec(z) + 1\right)^2} +3e^z\end{aligned}$$ (b) We have $p(0) = 0+3+1 = 4$ and $p'(0) = 0+3=3$, so an equation for the tangent line to $p$ at the point where $z=0$ is $y - 4 = 3(x-0)$. (c) We found that $p'(0) = 3$ is positive, so $p$ is increasing at $t=0$.