题目
1.设f(x)是偶函数,且f(0)存在,则 f'(0)=,2.设x=1+t^2;y=cost,. 则(d^2y)/(dx^2)= 3.设 y=ln(1+3^(-x)) ,则dy=4.设y=f(x+y),其中f具有二阶导数,且其一阶导数不等于1,则(d^2y)/(dx^2)= 5设函数 y=1/(2x+3),则 y^((n))(0)=)=6.设f(x)二阶可导且f(0)=0, f'(0)=1 , f''(0)=2 ,则lim_(x→0)(f(x)-x)/(x^2)= 7.设f(x)一阶可导且f(0)=2,又lim_(x→0)(xf(x)-ln(1+2x))/(x-ln(1+x))=4 ,则f'(0)= 8.设f(x)二阶可导且 f'(0)≠q0 ,又x=f(t)-π,;y=f(e^(2t)-1);. ,(dy)/(dx)|_(x=0)= 9.设 f(x)=x(x-1)(x+2)(x-3)⋯(x+100) ,则 f'(0)=10.设 f'(x)=f^2(x) 则 f'(x)=11.设 f'(x_0)=-1 ,则lim_(x→0)x/(f(x_0-2x)-f(x_0-x))= 12.设f(x)是偶函数,且f(x)存在,则 f'(x) 为函数(填“奇”或“偶”)13.设x+y=tany,则dy=14设 y=ln√((1-x)/(1+x^2)),则 y''(0)=15.设 y=(1+sinx)^x ,则dy|_(x=π)=16.曲线y=lnx的与直线x+y=1垂直的切线方程为17.已知曲线 f(x)=x^n 在点(1,1)处的切线与x轴的交点为 (ξ_n,0) ,则 lim_(n→∞)f(ξ_n)=18对数螺线 ρ=e^θ ,在点 (ρ,θ)=(e^(-π/(2)),π/(2))处法线的直角坐标方程为
1.设f(x)是偶函数,且f(0)存在,则 f'(0)=,2.设x=1+t^2;y=cost,. 则(d^2y)/(dx^2)= 3.设 y=ln(1+3^(-x)) ,则dy=4.设y=f(x+y),其中f具有二阶导数,且其一阶导数不等于1,则(d^2y)/(dx^2)= 5设函数 y=1/(2x+3),则 y^((n))(0)=)=6.设f(x)二阶可导且f(0)=0, f'(0)=1 , f''(0)=2 ,则lim_(x→0)(f(x)-x)/(x^2)= 7.设f(x)一阶可导且f(0)=2,又lim_(x→0)(xf(x)-ln(1+2x))/(x-ln(1+x))=4 ,则f'(0)= 8.设f(x)二阶可导且 f'(0)≠q0 ,又x=f(t)-π,;y=f(e^(2t)-1);. ,(dy)/(dx)|_(x=0)= 9.设 f(x)=x(x-1)(x+2)(x-3)⋯(x+100) ,则 f'(0)=10.设 f'(x)=f^2(x) 则 f'(x)=11.设 f'(x_0)=-1 ,则lim_(x→0)x/(f(x_0-2x)-f(x_0-x))= 12.设f(x)是偶函数,且f(x)存在,则 f'(x) 为函数(填“奇”或“偶”)13.设x+y=tany,则dy=14设 y=ln√((1-x)/(1+x^2)),则 y''(0)=15.设 y=(1+sinx)^x ,则dy|_(x=π)=16.曲线y=lnx的与直线x+y=1垂直的切线方程为17.已知曲线 f(x)=x^n 在点(1,1)处的切线与x轴的交点为 (ξ_n,0) ,则 lim_(n→∞)f(ξ_n)=18对数螺线 ρ=e^θ ,在点 (ρ,θ)=(e^(-π/(2)),π/(2))处法线的直角坐标方程为
题目解答
答案
1.02.(sint-tcost)/(4t^3) 3 dy=(-3^(-x)ln3)/(1+3^(-x))dx(f^n)/((1-f')^2) 5 (-1)^n(2/3)^n⋅1/3n!6.1 7.0 8.3 9. 100! 10. n!f^(n+1)(x)11.1.提示:因为lim_(x→0)(f(x_0-2x)-f(x_0-x))/x =lim_(x→0)(f(x_0-2x)-f(x_0))/x lim_(x→0)(f(x_0-x)-f(x_0^x))/x =-2f'(x_0)+f'(x_0)=-f'(x_0)=1 .12.奇13. cot2ydxx14 3/215.-πdx16.y=x-117.1/e .提示:曲线在(1,1)处的切线斜率k=f'(1)=n ,故切线方程为y-1=n(x-1).令y=0,得与x轴的交点 ξ_n=1-1/n ,所以lim_(n→∞)f(ξ_n)=lim_(n→∞)(1-1/n)^n=1/e 18. y-x=e^(π/(2)) .提示:将p=e化为直角坐标方√(x^2+y^2)=e^(arctany/x) ∴ln√(x^2+y^2)=arctany/x 两边对x求导1/2⋅(2x+2y+y^2)/(x^2+y^2)=1/(1+(y/2)^2)⋅(xy'-y)/(x^2)解之,得 x+yy'=xy'-y ,则 y'=(x+y)/(x-y)x-y在点 (ρ,θ)=(e^(π/(2)),π/(2)) 处处,有x_0=ρcosθ= =e^(π/(2))cosπ/(2)=0y_0=ρsinθ=e^(x/2)sinπ/2=e^(π/2),y'=-1 切线斜率 k_(vJ)=-1 ,法线斜率k法=1.法线方程 y-y_0=k_(ik)(x-x_0)即y-x=e.