합성함수의 미분법

(a) Since $h(x) = f(g(x))$, by the chain rule we know that $h'(x) = f'(g(x)) \cdot g'(x)$. Using the facts that $f(x)=x^3$ and $g(x) = \sin(x)$, it follows $f'(x) = 3x^2$ and $f'(g(x)) = 3\sin^2(x)$, and $g'(x) = \cos(x)$. Therefore, $$h'(x) = 3\sin^2(x) \cdot \cos(x)$$ and $h'\left( \frac{\pi}{4} \right) = 3 \left( \frac{1}{\sqrt{2}} \right)^2 \cdot \frac{1}{\sqrt{2}} = \frac{3}{2\sqrt{2}}$. (b) Note that $r'(x) = g'(f(x)) \cdot f'(x) = \cos(x^3) \cdot 3x^2$. Thus, we have $r'(0.25) = \cos(0.25^3) \cdot 3(0.25)^2 \approx 0.1875$, while using our work from (a), $h'(0.25) = 3\sin^2(0.25) \cdot \cos(0.25) \approx 0.1779$, so since $|h'(0.25) < r'(0.25)|$, we see that $r$ is changing more rapidly. (c) We have observed in (a) and (b) that $h'(x) = 3\sin^2(x) \cdot \cos(x)$ and $r'(x) = \cos(x^3) \cdot 3x^2$. Because both parts of the product in $h'(x)$ are periodic functions, $h'(x)$ is periodic (with period $2\pi$). However, $r'(x)$ is not periodic, since $3x^2$ grows without bound as $x$ increases.