113年中央地質碩士班-微積分詳解

 國立中央大學113學年度碩士班考試入學

所別：應用地質研究所
科目微積分

解答： $$\textbf{a.}\;f(x)=3x^4+5x^2-x+7 \Rightarrow f'(x)= \bbox[red, 2pt]{12x^3+10x-1}\\ \textbf{b.}\; f(x)= \frac{x^5+2x^2-6}{x^3} \Rightarrow f'(x)= \frac{5x^4+ 4x}{x^3}-{3(x^5+2x^2-6) \over x^4} = \bbox[red, 2pt]{2x^5 -2x^2+18 \over x^4}$$

解答： $$\textbf{a.}\;f(x)= \sin x-\cos x \Rightarrow f'(x)=\cos x+ \sin x \Rightarrow f''(x)= \bbox[red, 2pt]{-\sin x+\cos x} \\\textbf{b.}\; f(x) =3x^4+5x-6 \Rightarrow f'(x) =12x^3+5 \Rightarrow f''(x) = \bbox[red, 2pt]{36x^2}$$

解答： $$f(x) =\sin(\cos x) \Rightarrow f'(x)= \cos(\cos x) \cdot (-\sin x) = \bbox[red, 2pt]{-\sin x\cos(\cos x)}$$




解答： $$f(x) = \left({x+3\over x-2} \right)^3  = \left(1+{5\over x-2} \right)^3  \Rightarrow f'(x) = 3\left(1+{5\over x-2} \right)^2  \left(-{5\over (x-2)^2} \right) \\ \Rightarrow f'(1) =3(1-5)^2 (-5) = \bbox[red, 2pt]{-240}$$

解答： $$f(x^2) =x^3+5x \Rightarrow f'(x^2)\cdot (2x) =3x^2+5 \Rightarrow f'(x^2)={3x^2+5\over 2x}\\ \quad x^2=4 \Rightarrow x=2 \Rightarrow f'(4)={3\cdot 4+5\over 4} = \bbox[red, 2pt]{17\over 4}$$

解答：

$$\textbf{a.}\; y=\sqrt{1-x^2} \Rightarrow x^2+y^2=1 \Rightarrow \int_0^1 \sqrt{1-x^2}\,dx 為四分之一單位圓面積= \bbox[red,2pt]{\pi \over 4}$$

$$\textbf{b.}\;  \int_1^3 2x\,d x= \left. \left[ x^2 \right] \right|_1^3 =9-1= \bbox[red, 2pt] 8$$

$$\textbf{c.}\;  \int_1^3 2x\,d x= \left. \left[ x^2 \right] \right|_1^3 =9-1= \bbox[red, 2pt] 8$$

解答： $$\textbf{a.}\;  u=1+4x \Rightarrow du=4dx \Rightarrow \int \sqrt[3]{1+4x}\,dx = {1\over 4} \int u^{1/3}\,du ={3\over 16}u+c = \bbox[red, 2pt]{{3\over 16}(1+4x)+c}\\ \textbf{b.} \; u=x^3 \Rightarrow du=3x^2\,dx \Rightarrow \int x^2 \sin(x^3)\, dx = {1\over 3}\int \sin u\,du ={1\over 3}(-\cos u) +c = \bbox[red, 2pt]{-{1\over 3}\cos x^3+c}$$

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解題僅供參考，其他碩士班歷年考題