CLP-1 · 2.9절 · 25번우리말 번역도함수를 구하세요. y=e3x1+x2y=\frac{e^{3x}}{1+x^2}y=1+x2e3x 영어 원문 보기Evaluate the derivative. y=e3x1+x2y=\frac{e^{3x}}{1+x^2}y=1+x2e3x 풀이 공책 펴기한 줄씩 식 확인정답·풀이 보기연쇄 법칙에 의해 ddx{e3x}=3e3x\frac{\mathrm{d}}{\mathrm{d}x}\{e^{3x}\}=3e^{3x}dxd{e3x}=3e3x 임에 주의하여 몫의 미분법을 사용한다. y′=(1+x2)⋅3e3x−e3x(2x)(1+x2)2=e3x(3x2−2x+3)(1+x2)2\begin{aligned} y'&=\frac{(1+x^2)\cdot3e^{3x}-e^{3x}(2x)}{(1+x^2)^2}\\ &=\frac{e^{3x}(3x^2-2x+3)}{(1+x^2)^2} \end{aligned}y′=(1+x2)2(1+x2)⋅3e3x−e3x(2x)=(1+x2)2e3x(3x2−2x+3) 원본 교재 풀이(영어)We use the quotient rule, noting by the chain rule that ddx{e3x}=3e3x\frac{\mathrm{d}}{\mathrm{d}x}\{e^{3x}\}=3e^{3x}dxd{e3x}=3e3x: y′=(1+x2)⋅3e3x−e3x(2x)(1+x2)2=e3x(3x2−2x+3)(1+x2)2\begin{aligned} y'&=\frac{(1+x^2)\cdot3e^{3x}-e^{3x}(2x)}{(1+x^2)^2}\\ &=\frac{e^{3x}(3x^2-2x+3)}{(1+x^2)^2} \end{aligned}y′=(1+x2)2(1+x2)⋅3e3x−e3x(2x)=(1+x2)2e3x(3x2−2x+3) 한 번 더 — 비슷한 문제CLP-1 · 2.9절 · 24번우리말 번역도함수를 구하세요. y=(1+x2)cos2xy=(1+x^2)\cos^2 xy=(1+x2)cos2x 풀이 공책 펴기한 줄씩 식 확인다 풀었으면 정답 확인y=(1+x2)cos2xUsing the product rule,y′=(2x)cos2x+(1+x2)ddx{cos2x}Here, we’ll need to use the chain rule. Remember cos2x=[cosx]2.=2xcos2x+(1+x2)2cosx⋅ddx{cosx}=2xcos2x+(1+x2)2cosx⋅(−sinx)=2xcos2x−2(1+x2)sinxcosx\begin{aligned} y&=(1+x^2)\cos^2 x \\ \text{Using the product rule,} \\ y'&=(2x)\cos^2 x + (1+x^2)\frac{\mathrm{d}}{\mathrm{d}x}\{\cos^2 x\} \\ \text{Here, we'll need to use the chain rule. Remember $\cos^2 x = [\cos x]^2$.} \\ &=2x\cos^2x+(1+x^2) 2 \cos x \cdot \frac{\mathrm{d}}{\mathrm{d}x}\{ \cos x\}\\ &=2x\cos^2x+(1+x^2) 2\cos x \cdot (-\sin x)\\ &=2x\cos^2x-2(1+x^2) \sin x\cos x \end{aligned}yUsing the product rule,y′Here, we’ll need to use the chain rule. Remember cos2x=[cosx]2.=(1+x2)cos2x=(2x)cos2x+(1+x2)dxd{cos2x}=2xcos2x+(1+x2)2cosx⋅dxd{cosx}=2xcos2x+(1+x2)2cosx⋅(−sinx)=2xcos2x−2(1+x2)sinxcosx 이 문제로 이어서 풀기 # 합성함수의 미분법계산●●●보통자동 분류