高中倒數(shù)題目及答案

一、單項(xiàng)選擇題（每題2分，共10題）1.函數(shù)\(f(x)=x^{2}\)在\(x=1\)處的導(dǎo)數(shù)是（）A.0B.1C.2D.32.曲線\(y=\sinx\)在點(diǎn)\((\frac{\pi}{2},1)\)處的切線斜率為（）A.0B.1C.-1D.23.若\(f(x)=e^{x}\)，則\(f^\prime(0)\)等于（）A.0B.1C.eD.\(\frac{1}{e}\)4.函數(shù)\(y=\frac{1}{x}\)的導(dǎo)數(shù)是（）A.\(\frac{1}{x^{2}}\)B.\(-\frac{1}{x^{2}}\)C.\(\frac{1}{x}\)D.\(-\frac{1}{x}\)5.已知\(f(x)=x^{3}\)，\(f^\prime(x_0)=3\)，則\(x_0\)的值為（）A.1B.-1C.1或-1D.\(\sqrt{3}\)6.曲線\(y=x^{3}-2x+1\)在點(diǎn)\((1,0)\)處的切線方程為（）A.\(x-y-1=0\)B.\(x-y+1=0\)C.\(x+y-1=0\)D.\(x+y+1=0\)7.函數(shù)\(f(x)=x\cosx\)的導(dǎo)數(shù)是（）A.\(\cosx-x\sinx\)B.\(\cosx+x\sinx\)C.\(-\cosx+x\sinx\)D.\(-\cosx-x\sinx\)8.若\(f(x)=\lnx\)，則\(f^\prime(2)\)等于（）A.\(\frac{1}{2}\)B.2C.\(\ln2\)D.\(\frac{1}{\ln2}\)9.函數(shù)\(y=x^{2}+2x\)的導(dǎo)數(shù)在\(x=1\)處的值為（）A.4B.3C.2D.110.曲線\(y=\cosx\)在\(x=\frac{\pi}{4}\)處的切線斜率為（）A.\(\frac{\sqrt{2}}{2}\)B.\(-\frac{\sqrt{2}}{2}\)C.\(\frac{\sqrt{3}}{2}\)D.\(-\frac{\sqrt{3}}{2}\)答案：1.C2.A3.B4.B5.C6.A7.A8.A9.A10.B二、多項(xiàng)選擇題（每題2分，共10題）1.下列函數(shù)求導(dǎo)正確的是（）A.若\(y=x^{5}\)，則\(y^\prime=5x^{4}\)B.若\(y=\cosx\)，則\(y^\prime=-\sinx\)C.若\(y=\lnx\)，則\(y^\prime=\frac{1}{x}\)D.若\(y=e^{x}\)，則\(y^\prime=e^{x}\)2.曲線\(y=x^{3}\)在以下哪些點(diǎn)處的切線斜率為3（）A.\((1,1)\)B.\((-1,-1)\)C.\((\sqrt{3},3\sqrt{3})\)D.\((-\sqrt{3},-3\sqrt{3})\)3.函數(shù)\(f(x)=ax^{2}+bx+c\)（\(a\neq0\)）的導(dǎo)數(shù)為（）A.\(f^\prime(x)=2ax+b\)B.其導(dǎo)數(shù)是一次函數(shù)C.當(dāng)\(a\gt0\)時(shí)，\(f^\prime(x)\)是增函數(shù)D.當(dāng)\(a\lt0\)時(shí)，\(f^\prime(x)\)是減函數(shù)4.已知函數(shù)\(y=f(x)\)在點(diǎn)\((x_0,f(x_0))\)處的切線方程為\(y=2x+1\)，則（）A.\(f^\prime(x_0)=2\)B.\(f(x_0)=2x_0+1\)C.點(diǎn)\((x_0,f(x_0))\)在切線上D.切線斜率為25.下列關(guān)于導(dǎo)數(shù)的說(shuō)法正確的是（）A.導(dǎo)數(shù)表示函數(shù)在某一點(diǎn)處的變化率B.函數(shù)的導(dǎo)數(shù)大于0時(shí)，函數(shù)單調(diào)遞增C.函數(shù)的導(dǎo)數(shù)小于0時(shí)，函數(shù)單調(diào)遞減D.導(dǎo)數(shù)為0的點(diǎn)一定是函數(shù)的極值點(diǎn)6.函數(shù)\(y=x^{3}-3x\)的導(dǎo)數(shù)為（）A.\(y^\prime=3x^{2}-3\)B.令\(y^\prime=0\)，可得\(x=\pm1\)C.當(dāng)\(x\lt-1\)時(shí)，\(y^\prime\gt0\)，函數(shù)單調(diào)遞增D.當(dāng)\(-1\ltx\lt1\)時(shí)，\(y^\prime\lt0\)，函數(shù)單調(diào)遞減7.若\(f(x)=\sinx+\cosx\)，則（）A.\(f^\prime(x)=\cosx-\sinx\)B.\(f^\prime(\frac{\pi}{4})=0\)C.\(f(x)\)在\((0,\frac{\pi}{4})\)上單調(diào)遞增D.\(f(x)\)在\((\frac{\pi}{4},\frac{\pi}{2})\)上單調(diào)遞減8.函數(shù)\(y=\frac{1}{x^{2}}\)的導(dǎo)數(shù)相關(guān)正確的是（）A.可化為\(y=x^{-2}\)B.其導(dǎo)數(shù)\(y^\prime=-2x^{-3}=-\frac{2}{x^{3}}\)C.當(dāng)\(x\gt0\)時(shí)，\(y^\prime\lt0\)，函數(shù)單調(diào)遞減D.當(dāng)\(x\lt0\)時(shí)，\(y^\prime\gt0\)，函數(shù)單調(diào)遞增9.已知曲線\(y=x^{4}\)，則（）A.其導(dǎo)數(shù)\(y^\prime=4x^{3}\)B.在點(diǎn)\((1,1)\)處的切線斜率為4C.在點(diǎn)\((1,1)\)處的切線方程為\(y-1=4(x-1)\)D.該曲線的切線斜率恒大于等于010.對(duì)于函數(shù)\(f(x)=x\lnx\)，下列說(shuō)法正確的是（）A.其導(dǎo)數(shù)\(f^\prime(x)=\lnx+1\)B.\(f^\prime(1)=1\)C.當(dāng)\(0\ltx\lt\frac{1}{e}\)時(shí)，\(f^\prime(x)\lt0\)，函數(shù)單調(diào)遞減D.當(dāng)\(x\gt\frac{1}{e}\)時(shí)，\(f^\prime(x)\gt0\)，函數(shù)單調(diào)遞增答案：1.ABCD2.AB3.ABCD4.ABCD5.ABC6.ABCD7.ABCD8.ABCD9.ABC10.ABCD三、判斷題（每題2分，共10題）1.函數(shù)\(y=c\)（\(c\)為常數(shù)）的導(dǎo)數(shù)為0。（）2.曲線\(y=x^{2}\)在點(diǎn)\((0,0)\)處的切線方程是\(y=0\)。（）3.若\(f(x)=x^{n}\)（\(n\)為有理數(shù)），則\(f^\prime(x)=nx^{n-1}\)。（）4.函數(shù)\(y=\sinx\)的導(dǎo)數(shù)是\(y^\prime=\cosx\)，\(y=\cosx\)的導(dǎo)數(shù)是\(y^\prime=\sinx\)。（）5.函數(shù)\(f(x)\)在點(diǎn)\(x_0\)處的導(dǎo)數(shù)\(f^\prime(x_0)\)就是曲線\(y=f(x)\)在點(diǎn)\((x_0,f(x_0))\)處的切線斜率。（）6.若函數(shù)\(y=f(x)\)在區(qū)間\((a,b)\)內(nèi)\(f^\prime(x)\gt0\)，則\(y=f(x)\)在\((a,b)\)內(nèi)單調(diào)遞增。（）7.函數(shù)\(y=x^{3}\)在\(R\)上是單調(diào)遞增函數(shù)。（）8.函數(shù)\(f(x)=e^{x}\)與\(g(x)=\lnx\)的導(dǎo)數(shù)互為倒數(shù)。（）9.曲線\(y=\frac{1}{x}\)在點(diǎn)\((1,1)\)處的切線斜率為-1。（）10.函數(shù)\(y=x^{2}+1\)的導(dǎo)數(shù)在\(x=0\)處的值為0。（）答案：1.√2.√3.√4.×5.√6.√7.√8.×9.√10.√四、簡(jiǎn)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\((x^{n})^\prime=nx^{n-1}\)，常數(shù)的導(dǎo)數(shù)為0。\(y^\prime=(x^{3})^\prime-(4x)^\prime+(2)^\prime=3x^{2}-4\)。2.曲線\(y=x^{2}\)在點(diǎn)\((2,4)\)處的切線方程是什么？答案：先求導(dǎo)，\(y^\prime=2x\)，把\(x=2\)代入得切線斜率\(k=2×2=4\)。由點(diǎn)斜式得切線方程\(y-4=4(x-2)\)，即\(4x-y-4=0\)。3.函數(shù)\(f(x)=\lnx+x\)的單調(diào)遞增區(qū)間是？答案：求導(dǎo)得\(f^\prime(x)=\frac{1}{x}+1\)。令\(f^\prime(x)\gt0\)，即\(\frac{1}{x}+1\gt0\)，\(\frac{1+x}{x}\gt0\)，解得\(x\gt0\)。所以單調(diào)遞增區(qū)間是\((0,+\infty)\)。4.已知函數(shù)\(f(x)=x^{2}e^{x}\)，求\(f^\prime(x)\)。答案：根據(jù)乘積求導(dǎo)法則\((uv)^\prime=u^\primev+uv^\prime\)。\(u=x^{2}\)，\(u^\prime=2x\)；\(v=e^{x}\)，\(v^\prime=e^{x}\)。則\(f^\prime(x)=2xe^{x}+x^{2}e^{x}=e^{x}(x^{2}+2x)\)。五、討論題（每題5分，共4題）1.討論函數(shù)\(y=x^{3}-3x^{2}\)的單調(diào)性與極值情況。答案：求導(dǎo)得\(y^\prime=3x^{2}-6x=3x(x-2)\)。令\(y^\prime=0\)，得\(x=0\)或\(x=2\)。當(dāng)\(x\lt0\)或\(x\gt2\)時(shí)，\(y^\prime\gt0\)，函數(shù)遞增；當(dāng)\(0\ltx\lt2\)時(shí)，\(y^\prime\lt0\)，函數(shù)遞減。所以極大值\(y(0)=0\)，極小值\(y(2)=-4\)。2.探討曲線\(y=\sinx\)與\(y=\cosx\)在\([0,2\pi]\)上導(dǎo)數(shù)的幾何意義及變化情況。答案：\(y=\sinx\)導(dǎo)數(shù)\(y^\prime=\cosx\)，\(y=\cosx\)導(dǎo)數(shù)\(y^\prime=-\sinx\)。導(dǎo)數(shù)幾何意義是切線斜率。在\([0,2\pi]\)上，\(\cosx\)決定\(y=\sinx\)切線斜率變化，\(-\sinx\)決定\(y=\cosx\)切線斜率變化，如\(x=\frac{\pi}{2}\)時(shí)，\(y=\sinx\)切線斜率為0。3.分析函數(shù)\(f(x)=\frac{1}{x-1}\)的導(dǎo)數(shù)及在不同區(qū)間的單調(diào)性。答案：對(duì)\(f(x)=\frac{1}{x-1}=(x-1)^{-1}\)求導(dǎo)，\(f^\prime(x)=-\frac{1}{(x-1)^{2}}\)。\(f^\prime(x)\lt0\)恒成立，\(x\neq1\)。所以在\((-\inf