삼각함수의 미분

(a) Since $f(x) = \sin(x) \cot(x)$, by the product rule we know that $f'(x) = \sin(x) \cdot (-\csc^2(x)) + \cot(x) \cdot \cos(x)$. (b) False. Since $\cot(x) = \frac{\cos(x)}{\sin(x)}$, we can write $f(x) = \sin(x) \cdot \frac{\cos(x)}{\sin(x)}$. From here, it is tempting to “cancel” the $\sin(x)$ term in the numerator and denominator. However, note that because of the $\sin(x)$ in the denominator, $f$ is not defined for $x = \frac{\pi}{2} + k\pi$ for any integer $k$. In other words, the domain of $f$ is all real numbers except odd multiples of $\frac{\pi}{2}$. Thus, we can say that $f(x) = \cos(x)$ for all real numbers except odd multiples of $\frac{\pi}{2}$. But it is not the case that $f(x) = \cos(x)$ for every value of $x$. (c) Taking our result from (a) and converting $\csc(x)$ and $\cot(x)$ to expressions involving the sine and cosine functions, we have $$\begin{aligned}f'(x) &= \sin(x) \cdot (-\csc^2(x)) + \cot(x) \cdot \cos(x) \\ &= \sin(x) \cdot \left(-\frac{1}{\sin^2(x)}\right) + \frac{\cos(x)}{\sin(x)} \cdot \cos(x)\end{aligned}$$ Provided that $x \ne \frac{\pi}{2} + k\pi$ for some integer value of $k$, it follows that $$f'(x) = -\frac{1}{\sin(x)} + \frac{\cos^2(x)}{\sin(x)} = \frac{\cos^2(x)-1}{\sin(x)}\,.$$ By the Fundamental Trigonometric Identity, recall that $\cos^2(x) + \sin^2(x) = 1$, and thus $\cos^2(x)-1 = -\sin^2(x)$. Hence, for $x \ne \frac{\pi}{2} + k\pi$ for some integer value of $k$, we have $f'(x) = \frac{-\sin^2(x)}{\sin(x)} = -\sin(x)$. Again, this equality holds only for $x \ne \frac{\pi}{2} + k\pi$ for some integer value of $k$ because those values of $x$ are not in the domain of the original function $f$.