삼각함수의 미분

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

(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 $\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$.