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3.2.2 函数的和、差、积、商的导数基础达标1已知f(x)x33xln 3,则f(x)_.解析:f(x)(x3)(3x)(ln 3)3x23xln 303x23xln 3.答案:3x23xln 32设y2exsin x,则y_.解析:y2(ex)sin xex(sin x)2(exsin xexcos x)2ex(sin xcos x)答案:2ex(sin xcos x)3已知f(x)ax33x22,若f(1)4,则a的值是_解析:f(x)3ax26x,f(1)3a64,a.答案:4.已知a为实数,f(x)(x24)(xa),且f(1)0,则a_.,解析:f(x)(x24)(xa),x3ax24x4a,,f(x)3x22ax4.,又f(1)32a40,a.答案:5.函数y的导数是_解析:y.答案:6.设曲线y在点(3,2)处的切线与直线axy10垂直,则a_.解析:y1,y.曲线在点(3,2)处的切线斜率k.a2,即a2.答案:27求下列函数的导数:(1)y(2x23)(3x1);(2)y(2)2;(3)yxsincos;(4)y;(5)y2xcos x3xlog2 014x;(6)y.解:(1)法一:y(2x23)(3x1)(2x23)(3x1)4x(3x1)3(2x23)18x24x9.法二:y(2x23)(3x1)6x32x29x3,y(6x32x29x3)18x24x9.(2)y(2)2x44,yx(4)414x12x.(3)yxsincosxsin x,yx(sin x)1cos x.(4)y.(5)y(2x)cos x(cos x)2x3xlog2 014x(log2 014x)x2xln 2cos xsin x2x3log2 014x(log2 014e)x2xln 2cos x2xsin x3log2 014x3log2 104e.(6)ycos xsin x,ysin xcos x.8求过点(1,1)的曲线yx32x的切线方程解:设P(x0,y0)为切点,则切线的斜率为kf(x0)3x2,故切线方程为yy0(3x2)(xx0),即y(x2x0)(3x2)(xx0),又知切线过点(1,1),代入上述方程,得1(x2x0)(3x2)(1x0),解得x01或x0,故所求的切线方程为y1x1或y(x),即xy20或5x4y10.能力提升1若函数f(x)x3f(1)x2x5,则f(1)_.解析:f(x)x3f(1)x2x5,f(x)x22f(1)x1,将x1代入上式得f(1)12f(1)1,f(1)2,再令x1,得f(1)6.答案:62设函数f(x)g(x)x2,曲线yg(x)在点(1,g(1)处的切线方程为y2x1,则曲线yf(x)在点(1,f(1)处切线的斜率为_解析:依题意得f(x)g(x)2x,f(1)g(1)24.答案:43点P是曲线yex上任意一点,求点P到直线yx的最小距离解:根据题意设平行于直线yx的直线与曲线yex相切于点(x0,y0),该切点即为与yx距离最近的点,如图则在点(x0,y0)处的切线斜率为1.y(ex)ex,ex01,得x00,代入yex,得y01,即P(0,1)利用点到直线的距离公式得距离为.4设函数f(x)ax,曲线yf(x)在点(2,f(2)处的切线方程为7x4y120.(1)求f(x)的解析式;(2)证明:曲线yf(x)上任一点处的切线与直线x0和直线yx所围成的三角形的面积为定值,并求此定值解:(1)由7x4y120得yx3.当x2时,y,f(2),又f(x)a,f(2),由,得解之得.故f(x)x.(2)证明:设P(x0,y0)为曲线上任一点,由y1知,曲线在点P(x0,y0)处的切线方程为yy0(1)(xx0),即y(x0)(1)(xx0)令x0得y,从而得切线与直线x0的交点坐标为(0,)令yx得yx2x0,从而得切线与直线yx的交点坐标为(2x0,2x0)所以点P(x0,y0)处的切线与直线x0,yx所围成的三角形面积为|2x0|6.
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