삼각함수의 미분

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