基本初等函数的导数公式及导数运算法则综合测试题（附答案）选修2-21.2.2第2课时基本初等函数的导数公式及导数运算法则一、选择题1 .函数y = (x+ 1)2(x—1)在x= 1处的导数等于()A．1B．2C. 3D. 4答案］D解析］y = (x+1)2］'—x1 )+(x+ 1)2(x—1)'=2(x + 1)?(x—1) + (x+ 1)2= 3x2 + 2x—1,y‘ =1= 4.2.若对任意x€ R, f‘ =)4x3, f(1) = —1,则f(x)=()A. x4B. x4— 2C. 4x3—5D. x4+ 2答案］B解析］丁f‘(=4x3.f(x) = x4+c,又f(1) = — 1• • • 1 + c= — 1 ,• • • c= —2,—f(x) = x4 — 2.3 .设函数f(x) = xm + ax 的导数为f‘ =)2x+1,则数列{1f(n)}(n € N*) 的前n 项和是()A.nn+1B.n+2n+1C.nn—1D.n+1n 答案］A解析］T f(x) = xm+ ax 的导数为f‘(x)2x + 1,/. m = 2, a= 1,二f(x) = x2+ x,即f(n) = n2+n=n(n+ 1),二数列{1f(n)}(n € N*)的前n项和为：Sn= 11 X2 12X3 13 x+…+ 1n(n+ 1)=1 —12+ 12—13+…+ 1n —1n + 1=1 —1n+ 1= nn+ 1,故选 A.4.二次函数y = f(x)的图象过原点，且它的导函数y= f‘的)图象是过第一、二、三象限的一条直线，贝卩函数y= f(x)的图象的顶点在()A.第一象限B.第二象限C.第三象限D.第四象限答案］C解析］由题意可设f(x)= ax2 + bx, f' (=2ax + b,由于f‘(的图象是过第一、二、三象限的一条直线，故2a>0, b>0,则f(x) = ax+ b2a2—b24a, 顶点—b2a,—b24a 在第三象限,故选 C.5 .函数y = (2 + x3)2的导数为()A. 6x5+ 12x2B. 4+ 2x3C. 2(2+ x3)2D. 2(2+ x3)?3x答案］A解析］t y= (2+ x3)2= 4+ 4x3+ x6,/. y = 6x5 + 12x2.6. (2010?江西文，4)若函数f(x) = ax4 + bx2 + c满足f‘ 侍)2,贝卩 f -(1)=()A.- 1B.- 2C. 2D. 0答案］B解析］本题考查函数知识，求导运算及整体代换的思想，f‘(x)4ax3+ 2bx,f，41)=-4a-2b=- (4a + 2b), f '件)4a + 2b, A「—()= — f‘(4— 2要善于观察,故选 B.7.设函数f(x)= (1 —2x3)10，贝S f ' 4X)A. 0B.- 1C.- 60D. 60答案］D解析］f ' (4)10(1 —2x3)9(1 —2x3) 4 10(1 —2x3)9?(—6x2)= —60x2(1 —2X3)9,A f ' (1)60.8.函数y = sin2x—cos2x的导数是()A. 22cos2x— n 4B cos2x— sin2xC. sin2x+ cos2xD. 22cos2x+ n4答案］A解析］y = (si n2x—cos2x) = (sin 2x) —(cos2x)=2cos2x+ 2sin2x= 22cos2x— n 4.9.(2010?高二潍坊检测)已知曲线y= x24—3lnx的一条切线的斜率为12，则切点的横坐标为()A. 3B. 2C. 1D.12答案］A解析］由「(対x2 —3x= 12得x= 3.10.设函数f(x)是R上以5为周期的可导偶函数，则曲线y= f(x)在x=5 处的切线的斜率为()A.—15B. 0C.15D. 5答案］B解析］由题设可知f(x + 5) = f(x)二f‘ (+5)= f‘ ,)二f‘(®f‘ (0)又f( —x)= f(x),「. fTx)(—1)= f‘ (x)即x)= —f‘ ,)••• f‘ (0)0故f '(另f ' (&)0.故应选B.二、填空题11.________________________________________ 若f(x) = x, © (x 弄 1 + sin2x,则 f © (x并______________________ , © f(x)扫 _______答案］2si nx+ n4 1 + sin2x解析］f © (x)f 1 + sin2x= (sinx+ cosx)2 =|sinx + cosx| = 2sinx+ n 4.© f(x)” 1 + sin2x.12.设函数f(x) = cos(3x+ © )(0C ©< n,若f(x)+ f '是)奇函数，贝S ©=答案］n6解析］f (x)- 3sin(3x+ ©)f(x) + f ' (x)cos(3x+ © ) 3sin(3x+ ©)=2sin3x + ©+ 5 n 6.若f(x) + f‘ 为奇函数，则f(0) + f‘ (=)0,即0= 2sin + 5 n 6 二©+ 5 n 6= k n (l€ Z).又T ©€ (0, n ,二©= n 6.13.函数y= (1+ 2x2)8的导数为_________ .答案］32x(1 + 2x2)7解析］令u= 1 + 2x2,则y= u8,••• y' = y' u?u= 8u7?4x= 8(1 + 2x2)7?4x= 32x(1 + 2x2)7.14.函数y= x1 + x2 的导数为_______ .答案］(1 + 2x2)1 + x21 + x2解析］y = (x1 + x2) = x' + x2+ x(1 + x2) = 1 + x2 + x21 + x2= (1 + 2x2)1 + x21 + x2.三、解答题15.求下列函数的导数：(1)y= xsin2x; (2)y= In(x + 1+ x2);(3)y= ex+ lex—1; (4)y= x+ cosxx+ sinx.解析](1)y =(x) sir+xc(sin2x)'=sin2x+ x?2s in x?(s in x)= §in2x+ xsin 2x.(2)y = 1x+ 1 + x2?(x + 1 + x2)'=1x+ 1 + x2(1 + x1 + x2)= 11 + x2.(3)y = (ex +1) ' (—x1)—(ex + 1)(ex—1) ' (—1)2= —2ex(ex—1)2.(4)y = (x+ cosx) '+(sinx)—(x+ cosx)(x+ sinx)(x+ sinx)2= (1 —sinx)(x+ sinx)—(x+ cosx)(1+ cosx)(x+ sinx)2=—xcosx—xsinx+ sinx—cosx—1(x+ sinx)2.16．求下列函数的导数：(1)y= cos2(x2—x); (2)y= cosx?sin3x;(3)y= xIoga(x2+ x—1); (4)y= Iog2x—1x+ 1.解析](1)y = cos2(x2- x)]=2cos(x2— x)cos(x2— x)]=2cos(x2— x) —sin(x2—x)](x2 —x)'= 2cos(x2—x)—sin(x2—x)](2x—1)= (1 —2x)sin2(x2—x)．(2)y = (cosx?sin3x)= (cosx) ' s+3x)sx(sin3x) '=—sinxsin3x+ 3cosxcos3x= 3cosxcos3x—sinxsin3x.(3)y = Ioga(x2+ x—1)+ x?1x2+ x—1Iogae(x2+ x—1) = Ioga(x2+ x—1)+2x2＋xx2＋x－1logae.(4)y 厶x+ 1x—1x—1x+ 1' Iog2ex + 1x—1log2ex + 1 —x+ 1(x + 1)2=2log2ex2— 1.17.设f(x) = 2sinx1 + x2,如果 f '閑2(1 + x2)2?g(x),求g(x).解析］•/ f'閑2cosx(1+ x2)—2sinx?2x(1 + x2)2=2(1 + x2)2(1 + x2)cosx— 2x?s inx］,又f‘ 閑2(1 + x2)2?g(x).g(x)= (1 + x2)cosx- 2xs in x.18.求下列函数的导数：(其中f(x)是可导函数)(1)y= f1x；(2)y=f(x2 + 1).解析］(1)解法1：设y= f(u), u= 1x,则y‘亲y‘ u?u=f' (u—1x2= —1x2f ' 1x.解法2：y = f1x = f‘ 1x?似-1x2f‘ 1x.(2)解法1:设y= f(u), u = v, v=x2+ 1,。