高中導(dǎo)數(shù)相關(guān)試題及答案

一、單項選擇題（每題2分，共10題）1.函數(shù)\(y=x^2\)的導(dǎo)數(shù)是（）A.\(2x\)B.\(x^2\)C.\(2\)D.\(0\)2.若\(f(x)=e^x\)，則\(f^\prime(x)\)等于（）A.\(e^x\)B.\(-e^x\)C.\(xe^x\)D.\(\frac{1}{e^x}\)3.函數(shù)\(y=\sinx\)的導(dǎo)數(shù)是（）A.\(\cosx\)B.\(-\cosx\)C.\(\sinx\)D.\(-\sinx\)4.曲線\(y=x^3\)在點\((1,1)\)處的切線斜率為（）A.\(1\)B.\(2\)C.\(3\)D.\(4\)5.函數(shù)\(f(x)=x^3-3x\)的導(dǎo)數(shù)\(f^\prime(x)\)為（）A.\(3x^2-3\)B.\(x^2-3\)C.\(3x^2\)D.\(3x-3\)6.若\(y=\lnx\)，則\(y^\prime\)等于（）A.\(\frac{1}{x}\)B.\(x\)C.\(-\frac{1}{x}\)D.\(x^2\)7.函數(shù)\(y=2x^2+3x+1\)在\(x=1\)處的導(dǎo)數(shù)為（）A.\(4\)B.\(7\)C.\(5\)D.\(6\)8.已知\(f(x)=x^n\)，\(f^\prime(1)=2\)，則\(n\)的值為（）A.\(1\)B.\(2\)C.\(3\)D.\(4\)9.函數(shù)\(y=\cos2x\)的導(dǎo)數(shù)是（）A.\(-2\sin2x\)B.\(2\sin2x\)C.\(-\sin2x\)D.\(\sin2x\)10.曲線\(y=\frac{1}{x}\)在點\((1,1)\)處的切線方程是（）A.\(x+y-2=0\)B.\(x-y-2=0\)C.\(x+y+2=0\)D.\(x-y+2=0\)二、多項選擇題（每題2分，共10題）1.以下哪些函數(shù)的導(dǎo)數(shù)是\(2x\)（）A.\(x^2+1\)B.\(x^2-2\)C.\(2x^2\)D.\(x^2+C\)（\(C\)為常數(shù)）2.關(guān)于函數(shù)\(f(x)=x^3\)，下列說法正確的是（）A.\(f^\prime(x)=3x^2\)B.函數(shù)在\(R\)上單調(diào)遞增C.曲線\(y=f(x)\)在點\((0,0)\)處切線斜率為\(0\)D.\(f(x)\)是奇函數(shù)3.下列函數(shù)求導(dǎo)正確的是（）A.\((\cosx)^\prime=-\sinx\)B.\((e^{2x})^\prime=2e^{2x}\)C.\((\ln2x)^\prime=\frac{1}{x}\)D.\((x^{-2})^\prime=-2x^{-3}\)4.曲線\(y=x^3-3x^2+2\)在哪些點處切線斜率為\(0\)（）A.\((0,2)\)B.\((1,0)\)C.\((2,-2)\)D.\((3,2)\)5.函數(shù)\(f(x)=x^2e^x\)的導(dǎo)數(shù)\(f^\prime(x)\)可以是（）A.\(2xe^x\)B.\(e^x(x^2+2x)\)C.\(x(x+2)e^x\)D.\(x^2e^x+2xe^x\)6.若函數(shù)\(y=f(x)\)在某點處導(dǎo)數(shù)為\(0\)，則（）A.該點可能是函數(shù)的極值點B.函數(shù)在該點一定有最值C.函數(shù)在該點切線斜率為\(0\)D.函數(shù)在該點附近單調(diào)性可能改變7.已知\(f(x)\)可導(dǎo)，下列正確的是（）A.\((f(x)+C)^\prime=f^\prime(x)\)（\(C\)為常數(shù)）B.\((Cf(x))^\prime=Cf^\prime(x)\)C.\((f(x)g(x))^\prime=f^\prime(x)g(x)+f(x)g^\prime(x)\)D.\((\frac{f(x)}{g(x)})^\prime=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g^2(x)}\)（\(g(x)\neq0\)）8.函數(shù)\(y=\sinx+\cosx\)的導(dǎo)數(shù)\(y^\prime\)是（）A.\(\cosx-\sinx\)B.\(\sinx-\cosx\)C.\(-\sinx-\cosx\)D.\(\cosx+\sinx\)9.曲線\(y=\lnx\)上切線斜率為\(1\)的點有（）A.\((1,0)\)B.\((e,1)\)C.\((2,\ln2)\)D.\((e^2,2)\)10.下列函數(shù)中，在\((0,+\infty)\)上單調(diào)遞增的是（）A.\(y=x^2\)B.\(y=e^x\)C.\(y=\lnx\)D.\(y=\frac{1}{x}\)三、判斷題（每題2分，共10題）1.函數(shù)\(y=5\)的導(dǎo)數(shù)是\(5\)。（）2.若\(f(x)\)在\(x=a\)處可導(dǎo)，則曲線\(y=f(x)\)在點\((a,f(a))\)處有切線。（）3.函數(shù)\(y=x^4\)的導(dǎo)數(shù)比\(y=x^3\)的導(dǎo)數(shù)大。（）4.\(f^\prime(x_0)\)表示曲線\(y=f(x)\)在點\((x_0,f(x_0))\)處切線的斜率。（）5.函數(shù)\(y=\sin2x\)的導(dǎo)數(shù)是\(2\cos2x\)。（）6.若函數(shù)\(f(x)\)在區(qū)間\((a,b)\)內(nèi)\(f^\prime(x)>0\)，則\(f(x)\)在\((a,b)\)上單調(diào)遞增。（）7.函數(shù)\(y=x^2\)與\(y=x^2+1\)的導(dǎo)數(shù)相同。（）8.曲線\(y=\frac{1}{x}\)的導(dǎo)數(shù)恒小于\(0\)。（）9.函數(shù)\(f(x)=x^3\)在\(x=0\)處取得極值。（）10.若\(f^\prime(x)\)是偶函數(shù)，則\(f(x)\)是奇函數(shù)。（）四、簡答題（每題5分，共4題）1.求函數(shù)\(y=x^3-4x+2\)的導(dǎo)數(shù)。答案：根據(jù)求導(dǎo)公式\((x^n)^\prime=nx^{n-1}\)，\(y^\prime=3x^2-4\)。2.曲線\(y=x^2\)在點\((2,4)\)處的切線方程是什么？答案：先求導(dǎo)數(shù)\(y^\prime=2x\)，\(x=2\)時，切線斜率\(k=4\)。由點斜式得切線方程\(y-4=4(x-2)\)，即\(4x-y-4=0\)。3.求函數(shù)\(f(x)=e^x\cosx\)的導(dǎo)數(shù)。答案：根據(jù)乘積求導(dǎo)法則\((uv)^\prime=u^\primev+uv^\prime\)，\(u=e^x\)，\(u^\prime=e^x\)，\(v=\cosx\)，\(v^\prime=-\sinx\)，則\(f^\prime(x)=e^x\cosx-e^x\sinx=e^x(\cosx-\sinx)\)。4.已知函數(shù)\(y=\ln(x^2+1)\)，求其導(dǎo)數(shù)。答案：令\(u=x^2+1\)，則\(y=\lnu\)。先對\(y\)關(guān)于\(u\)求導(dǎo)得\(\frac{1}{u}\)，再對\(u\)關(guān)于\(x\)求導(dǎo)得\(2x\)，根據(jù)復(fù)合函數(shù)求導(dǎo)法則\(y^\prime=\frac{2x}{x^2+1}\)。五、討論題（每題5分，共4題）1.討論函數(shù)\(y=x^3-3x\)的單調(diào)性與極值情況。答案：求導(dǎo)得\(y^\prime=3x^2-3=3(x+1)(x-1)\)。令\(y^\prime=0\)，得\(x=\pm1\)。當(dāng)\(x<-1\)或\(x>1\)時，\(y^\prime>0\)，函數(shù)遞增；當(dāng)\(-1<x<1\)時，\(y^\prime<0\)，函數(shù)遞減。極大值為\(y(-1)=2\)，極小值為\(y(1)=-2\)。2.對于函數(shù)\(f(x)=x^2e^{-x}\)，討論其在\(R\)上的最值情況。答案：求導(dǎo)\(f^\prime(x)=2xe^{-x}-x^2e^{-x}=x(2-x)e^{-x}\)。令\(f^\prime(x)=0\)，得\(x=0\)或\(x=2\)。當(dāng)\(x<0\)或\(x>2\)時，\(f^\prime(x)<0\)，函數(shù)遞減；當(dāng)\(0<x<2\)時，\(f^\prime(x)>0\)，函數(shù)遞增。\(f(0)=0\)，\(f(2)=\frac{4}{e^2}\)，\(x\to\pm\infty\)時，\(f(x)\to0\)，所以最大值為\(\frac{4}{e^2}\)，最小值為\(0\)。3.若函數(shù)\(f(x)=\frac{1}{3}x^3+ax^2+bx\)在\(x=1\)處取得極值\(-\frac{4}{3}\)，討論\(a\)，\(b\)的值及函數(shù)單調(diào)性。答案：\(f^\prime(x)=x^2+2ax+b\)，由已知\(f(1)=-\frac{4}{3}\)，\(f^\prime(1)=0\)，即\(\begin{cases}\frac{1}{3}+a+b=-\frac{4}{3}\\1+2a+b=0\end{cases}\)，解得\(a=1\)，\(b=-3\)。\(f^\prime(x)=x^2+2x-3=(x+3)(x-1)\)。當(dāng)\(x<-3\)或\(x>1\)時，\(f^\prime(x)>0\)，函數(shù)遞增；當(dāng)\(-3<x<1\)時，\(f^\prime(x)<0\)，函數(shù)遞減。4.討論曲線\(y=\lnx\)與直線\(y=kx\)的交點情況。答案：令\(g(x)=\lnx-kx\)，\(g^\prime(x)=\frac{1}{x}-k\)。當(dāng)\(k\leq0\)時，\(g^\prime(x)>0\)，\(g(x)\)在\((0,+\infty)\)遞增，\(x\to0\)時，\(g(x)\to-\infty\)，\(x\to+\infty\)時，\(g(x)\to+\infty\)，有一個交點。當(dāng)\(k>0\)時，\(g(x)\)在\((0,\frac{1}{k})\)遞增，在\((\frac{1}{k},+\infty)\)遞減，\(g(x)_{max}=g(\frac{1}{k})=-\lnk-1\)。\(g(x)_{max}<0\)即\(k>\frac{1}{e}\)時無交點；\(k=\frac{1}{e}\)時有一個交