삼각함수의 미분

The graph $f(x)=\sin x$ has maximum slope at those points where $\cos x$ has a maximum. This makes sense, because $f'(x)=\cos x$: the maximum values of the slope of sine correspond to the maximum values of cosine.

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 $f'(\theta)=\frac{\mathrm{d}}{\mathrm{d}\theta}\{\cos \theta\}=-\sin(\theta)$.

We know the derivative of $\sin x$, but not of $\sin(-x)$. So we re-write $f(x)$ using identities:

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

We apply the quotient rule.

$$\begin{aligned} s'(\theta)&=\frac{(\cos \theta-\sin\theta)(-\sin\theta+\cos\theta)-(\cos\theta+\sin\theta)(-\sin\theta-\cos\theta)}{(\cos\theta-\sin\theta)^2}\\ &=\frac{(\cos \theta-\sin\theta)^2+(\cos\theta+\sin\theta)^2}{(\cos\theta-\sin\theta)^2}\\ &=1+\left(\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)^2 \end{aligned}$$

In order for $f$ to be differentiable at $x=0$, it must also be continuous at $x=0$. This forces $$\lim_{x\to 0^-}f(x) = \lim_{x\to 0^+}f(x) =f(0)\qquad\text{or}\qquad \lim_{x\to 0^-}\cos(x) = \lim_{x\to 0^+}(ax+b) =1$$ or $b=1$. In order for $f$ to be differentiable at $x=0$, we need the limit $$\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$$ to exist. This is the case if and only if the two one–sided limits

$$\begin{aligned} \lim_{h\to 0^-}\frac{f(0+h)-f(0)}{h} &=\lim_{h\to 0^-}\frac{\cos(h)-\cos(0)}{h} \\ \text{and} \\ \lim_{h\to 0^+}\frac{f(0+h)-f(0)}{h} &=\lim_{h\to 0^+}\frac{(ah+b)-\cos(0)}{h} =a\qquad\text{since $b=1$} \\ \text{exist and are equal. Because $\cos(x)$ is differentiable at $x=0$ we have} \\ \lim_{h\to 0^-}\frac{\cos(h)-\cos(0)}{h} &= \frac{\mathrm{d}}{\mathrm{d}x}\cos(x)\bigg|_{x=0} = -\sin(x)\Big|_{x=0}=0 \end{aligned}$$

So, we need $a=0$ and $b=1$.

We compute the derivative of $\cos(x)+2x$ as being $-\sin(x)+2$, which evaluated at $x=\frac{\pi}{2}$ yields $-1+2=1$. Since we also compute $\cos(\pi/2)+2(\pi/2)=0+\pi$, then the equation of the tangent line is

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

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 $$a=\frac{-6\sin 0(2+\sin 0+\cos 0)-6\cos 0(\cos 0-\sin 0)}{(2+\sin 0+\cos 0)^2} =\frac{-6}{(2+1)^2}\implies\boxed{a=-\frac{2}{3}}$$