삼각함수의 미분

$\frac{dy}{dx}=\frac{2{\csc }^{2}x}{{(1+\cot x)}^{2}}$

$3\cos x-x\sin x$

$\left(\frac{\pi }{4}, 1\right), \left(\frac{3\pi }{4}, -1\right), \left(\frac{5\pi }{4}, 1\right), \left(\frac{7\pi }{4}, -1\right)$

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 $b=1$.