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문제 은행 · 미적분 · 미적분Ⅰ · 미분

다항함수의 미분법

xⁿ의 미분, 상수배·합·차·곱의 미분법으로 다항함수의 도함수를 구한다

유형전체개념 확인계산그래프·표 해석활용증명·논술
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풀이전체우리말 풀이 있는 문제만

유형·난이도는 우리말 풀이가 붙은 문제는 선생님이 정했고, 나머지는 자동 분류라 다를 수 있어요. 난이도가 정해지지 않은 문제는 난이도로 거를 때 빠집니다.

1문제 · 1 / 1쪽

1.CLP-1 · 2.4절 · 17번우리말 번역

f(x)=g(x)h(x)f(x)=\dfrac{g(x)}{h(x)} 를 몫의 미분법으로 미분한 결과가, f(x)=g(x)k(x)⋅k(x)h(x)f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)} 를 곱의 미분법과 몫의 미분법으로 미분한 결과와 같음을 보이세요.

영어 원문 보기

Verify that differentiating f(x)=g(x)h(x)f(x)=\dfrac{g(x)}{h(x)} using the quotient rule gives the same answer as differentiating f(x)=g(x)k(x)⋅k(x)h(x)f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)} using the product rule and the quotient rule.

정답·풀이 보기

First expression, f(x)=g(x)h(x):f′(x)=h(x)g′(x)−g(x)h′(x)h2(x)Second expresson, f(x)=g(x)k(x)⋅k(x)h(x):f′(x)=(k(x)g′(x)−g(x)k′(x)k2(x))(k(x)h(x))+(g(x)k(x))(h(x)k′(x)−k(x)h′(x)h2(x))=k(x)g′(x)−g(x)k′(x)k(x)h(x)+g(x)h(x)k′(x)−g(x)k(x)h′(x)k(x)h2(x)=h(x)k(x)g′(x)−h(x)g(x)k′(x)k(x)h2(x)+g(x)h(x)k′(x)−g(x)k(x)h′(x)k(x)h2(x)=h(x)k(x)g′(x)−h(x)g(x)k′(x)+g(x)h(x)k′(x)−g(x)k(x)h′(x)k(x)h2(x)=h(x)k(x)g′(x)−g(x)k(x)h′(x)k(x)h2(x)=h(x)g′(x)−g(x)h′(x)h2(x)and this is exactly what we got from differentiating the first expression.\begin{aligned} \\ \text{First expression, $f(x)=\dfrac{g(x)}{h(x)}$:} \\ f'(x)&=\frac{h(x)g'(x)-g(x)h'(x)}{h^2(x)} \\ \text{Second expresson, $f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)}$:} \\ f'(x)&=\left(\frac{k(x)g'(x)-g(x)k'(x)}{k^2(x)}\right)\left(\frac{k(x)}{h(x)}\right)+\left(\frac{g(x)}{k(x)}\right)\left(\frac{h(x)k'(x)-k(x)h'(x)}{h^2(x)}\right)\\ &=\frac{k(x)g'(x)-g(x)k'(x)}{k(x)h(x)}+ \frac{g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)k(x)g'(x)-h(x)g(x)k'(x)}{k(x)h^2(x)}+ \frac{g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)k(x)g'(x)-h(x)g(x)k'(x)+g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)k(x)g'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)g'(x)-g(x)h'(x)}{h^2(x)} \\ \text{and this is exactly what we got from differentiating the first expression.} \\ \end{aligned}

원본 교재 풀이(영어)

First expression, f(x)=g(x)h(x):f′(x)=h(x)g′(x)−g(x)h′(x)h2(x)Second expresson, f(x)=g(x)k(x)⋅k(x)h(x):f′(x)=(k(x)g′(x)−g(x)k′(x)k2(x))(k(x)h(x))+(g(x)k(x))(h(x)k′(x)−k(x)h′(x)h2(x))=k(x)g′(x)−g(x)k′(x)k(x)h(x)+g(x)h(x)k′(x)−g(x)k(x)h′(x)k(x)h2(x)=h(x)k(x)g′(x)−h(x)g(x)k′(x)k(x)h2(x)+g(x)h(x)k′(x)−g(x)k(x)h′(x)k(x)h2(x)=h(x)k(x)g′(x)−h(x)g(x)k′(x)+g(x)h(x)k′(x)−g(x)k(x)h′(x)k(x)h2(x)=h(x)k(x)g′(x)−g(x)k(x)h′(x)k(x)h2(x)=h(x)g′(x)−g(x)h′(x)h2(x)and this is exactly what we got from differentiating the first expression.\begin{aligned} \\ \text{First expression, $f(x)=\dfrac{g(x)}{h(x)}$:} \\ f'(x)&=\frac{h(x)g'(x)-g(x)h'(x)}{h^2(x)} \\ \text{Second expresson, $f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)}$:} \\ f'(x)&=\left(\frac{k(x)g'(x)-g(x)k'(x)}{k^2(x)}\right)\left(\frac{k(x)}{h(x)}\right)+\left(\frac{g(x)}{k(x)}\right)\left(\frac{h(x)k'(x)-k(x)h'(x)}{h^2(x)}\right)\\ &=\frac{k(x)g'(x)-g(x)k'(x)}{k(x)h(x)}+ \frac{g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)k(x)g'(x)-h(x)g(x)k'(x)}{k(x)h^2(x)}+ \frac{g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)k(x)g'(x)-h(x)g(x)k'(x)+g(x)h(x)k'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)k(x)g'(x)-g(x)k(x)h'(x)}{k(x)h^2(x)}\\ &=\frac{h(x)g'(x)-g(x)h'(x)}{h^2(x)} \\ \text{and this is exactly what we got from differentiating the first expression.} \\ \end{aligned}

한 번 더 — 비슷한 문제 풀기CLP-1 · 2.4절 · 4번
# 다항함수의 미분법증명·논술도전자동 분류