第二章总复习题课

一、填空题

1.函数\begin{aligned} y=\frac{{\rm e}^x}{x} \end{aligned}在点\begin{aligned} x_0=2 \end{aligned}的微分为\begin{aligned} [\ \frac{{\rm e}^2}4{\rm d}x\ ]. \end{aligned}

解析：\begin{aligned} \left.{\rm d}y \right|_{x=2}&=\left.\left(\frac{{\rm e}^x}{x}\right)'\right|_{x=2}{\rm d}x\\ &=\left.\left(\frac{x{\rm e}^x-{\rm e}^x}{x^2}\right)\right|_{x=2}{\rm d}x\\ &=\frac{{\rm e}^2}4{\rm d}x \end{aligned}

2.函数\begin{aligned} y=x\ln x \end{aligned}在点\begin{aligned} x_0={\rm e} \end{aligned}的微分为[\ 2{\rm d}x\ ].

解析：\begin{aligned} y&=x\ln x,\left.{\rm d}x\right|_{x=e}\\ &=\left.(x\ln x)'\right|_{x=e}{\rm d}x\\ &=\left.(1+\ln x)\right|_{x=e}{\rm d}x\\ &=2{\rm d}x \end{aligned}

3.函数\begin{aligned} y={\rm e}^{-3x^2} \end{aligned}的导数\begin{aligned} y'=[\ -6x{\rm e}^{-3x^2}\ ]. \end{aligned}

4.函数\begin{aligned} y=\ln ({1+x^2}) \end{aligned}的导数\begin{aligned} y'=[\ \frac{2x}{1+x^2}\ ]. \end{aligned}

二、选择题

5.求函数\begin{aligned} y=x^{\frac1x} \end{aligned} (x>0)的导数\begin{aligned} y'=[\ A \ ]. \end{aligned}

A.\begin{aligned} x^{\frac1x-2}(1-\ln x) \end{aligned}

B.\begin{aligned} x^{\frac1x}(1-\ln x) \end{aligned}

C.\begin{aligned} x^{\frac1x-2}\ln x \end{aligned}

D.\begin{aligned} x^{\frac1x}\ln x \end{aligned}

6.求函数\begin{aligned} y=x^{\sin x}(x>0) \end{aligned}的导数y'=[\ B\ ].

A.\begin{aligned} x^{\sin x}\left(\sin x\ln x+\frac{\sin x}{x}\right) \end{aligned}

B.\begin{aligned} x^{\sin x}\left(\cos x\ln x+\frac{\sin x}{x}\right) \end{aligned}

C.\begin{aligned} x^{\sin x}\left(\cos x\ln x+\frac{\cos x}{x}\right) \end{aligned}

D.\begin{aligned} x^{\sin x}\left(\cos x\ln x-\frac{\sin x}{x}\right) \end{aligned}

解析：\begin{aligned} &\ln y=\ln x^{\sin x}=\sin x\ln x\\ &\frac1yy'=\cos x\ln x+\frac{\sin x}{x}\\ &y'=y\left(\cos x\ln x+\frac{\sin x}{x}\right)\\ &y'=x^{\sin x}\left(\cos x\ln x+\frac{\sin x}{x}\right) \end{aligned}

7.已知\begin{aligned} \lim_{\Delta x\to0}\frac{f(1-\Delta x)-f(1)}{\Delta x}=2 \end{aligned}，\begin{aligned} f'(1)=[\ C\ ]. \end{aligned}

A.1

B.2

C.-2

D.-1

解析：\begin{aligned} f'(1)&=\lim_{\Delta x\to0}\frac{f(1-\Delta x)-f(1)}{-\Delta x}\\ &=-\lim_{\Delta x\to 0}\frac{f(1-\Delta x)-f(1)}{\Delta x}\\ &=-2 \end{aligned}

8.已知f'(1)=2，\begin{aligned} \lim_{x\to0}\frac{f(1-2x)-f(1)}{x}=[\ D\ ]. \end{aligned}

A.4

B.2

C.-2

D.-4

解析：\begin{aligned} &\lim_{x\to0}\frac{f(1-2x)-f(1)}{x}\\ &=-2\lim_{x\to0}\frac{f(1-2x)-f(1)}{-2x}\\ &=-4 \end{aligned}

三、解答题

9.曲线方程为\begin{cases} x=2+t\\ y=t^2+2\sin t \end{cases}，求此曲线在x=2处的切线方程。

解答：t=0,x=2,y=0

\begin{aligned} \frac{{\rm d}y}{{\rm d}x}&=\frac{\frac{{\rm d}y}{{\rm d}t}}{\frac{{\rm d}x}{{\rm d}t}}\\ &=\frac{(t^2+2\sin t)'}{(2+t)'}\\ &=\frac {2t+2\cos t}{1}\\ &=2t+2\cos t\\ \end{aligned}

\begin{aligned} k=\left.\frac{{\rm d}y}{{\rm d}x}\right|_{t=0} &=\left.(2t+2\cos t)\right|_{t=0}=2 \end{aligned}

所求的切线方程为y=2(x-2)

10.求曲线\begin{cases} x=1+t^2\\ y=t^3\\ \end{cases}在t=2处的切线方程。

解答：t=2,x=5,y=8

\begin{aligned} \frac{{\rm d}y}{{\rm d}x}&=\frac{\frac{{\rm d}y}{{\rm d}t}}{\frac{{\rm d}x}{{\rm d}t}}\\ &=\frac{(t^3)'}{(1+t^2)'}\\ &=\frac{3t^2}{2t}\\ &=\frac32t\\ \end{aligned}

\begin{aligned} k=\left.\frac{{\rm d}y}{{\rm d}x}\right|_{t=2} &=\left.\frac{{\rm d}y}{{\rm d}x}\right|_{t=2}=3 \end{aligned}

所求的切线方程为y-8=3(x-5)

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