2025年專升本高等數(shù)學(xué)（二）模擬統(tǒng)考卷：多元函數(shù)微分學(xué)重點(diǎn)難點(diǎn)解析

2025年專升本高等數(shù)學(xué)（二）模擬統(tǒng)考卷：多元函數(shù)微分學(xué)重點(diǎn)難點(diǎn)解析一、選擇題（本大題共10小題，每小題5分，共50分）1.設(shè)函數(shù)f(x,y)=x^2y-y^2x，則f(x,y)在點(diǎn)(1,1)處的偏導(dǎo)數(shù)f_x'(1,1)為：A.2B.-2C.0D.12.設(shè)函數(shù)f(x,y)=e^x*y^2，則f(x,y)在點(diǎn)(0,1)處的偏導(dǎo)數(shù)f_x'(0,1)為：A.1B.2C.eD.e^23.設(shè)函數(shù)f(x,y)=x^3+y^3，則f(x,y)在點(diǎn)(0,0)處的全微分df(0,0)為：A.0B.dx+dyC.2dx+2dyD.3dx+3dy4.設(shè)函數(shù)f(x,y)=ln(x^2+y^2)，則f(x,y)在點(diǎn)(1,0)處的全微分df(1,0)為：A.0B.dx-dyC.dx+dyD.2dx-2dy5.設(shè)函數(shù)f(x,y)=arctan(x/y)，則f(x,y)在點(diǎn)(1,1)處的偏導(dǎo)數(shù)f_x'(1,1)為：A.0B.1C.-1D.無定義6.設(shè)函數(shù)f(x,y)=e^(x^2+y^2)，則f(x,y)在點(diǎn)(0,0)處的全微分df(0,0)為：A.0B.dx+dyC.2dx+2dyD.4dx+4dy7.設(shè)函數(shù)f(x,y)=sin(x^2+y^2)，則f(x,y)在點(diǎn)(0,0)處的全微分df(0,0)為：A.0B.dx+dyC.2dx+2dyD.4dx+4dy8.設(shè)函數(shù)f(x,y)=ln(x^2-y^2)，則f(x,y)在點(diǎn)(1,0)處的全微分df(1,0)為：A.0B.dx-dyC.dx+dyD.2dx-2dy9.設(shè)函數(shù)f(x,y)=arccos(x^2+y^2)，則f(x,y)在點(diǎn)(1,0)處的偏導(dǎo)數(shù)f_x'(1,0)為：A.0B.1C.-1D.無定義10.設(shè)函數(shù)f(x,y)=x^3*y^3，則f(x,y)在點(diǎn)(0,0)處的全微分df(0,0)為：A.0B.dx+dyC.2dx+2dyD.3dx+3dy二、填空題（本大題共5小題，每小題10分，共50分）1.設(shè)函數(shù)f(x,y)=x^2+y^2，則f(x,y)在點(diǎn)(1,1)處的全微分df(1,1)為______。2.設(shè)函數(shù)f(x,y)=e^(x^2+y^2)，則f(x,y)在點(diǎn)(0,0)處的偏導(dǎo)數(shù)f_x'(0,0)為______。3.設(shè)函數(shù)f(x,y)=arctan(x/y)，則f(x,y)在點(diǎn)(1,1)處的偏導(dǎo)數(shù)f_y'(1,1)為______。4.設(shè)函數(shù)f(x,y)=x^2y-y^2x，則f(x,y)在點(diǎn)(1,1)處的全微分df(1,1)為______。5.設(shè)函數(shù)f(x,y)=ln(x^2+y^2)，則f(x,y)在點(diǎn)(1,0)處的全微分df(1,0)為______。三、解答題（本大題共5小題，每小題20分，共100分）1.求函數(shù)f(x,y)=x^2+y^2在點(diǎn)(2,3)處的全微分。2.設(shè)函數(shù)f(x,y)=e^(x^2+y^2)，求f(x,y)在點(diǎn)(0,0)處的全微分。3.設(shè)函數(shù)f(x,y)=x^2y-y^2x，求f(x,y)在點(diǎn)(1,1)處的全微分。4.設(shè)函數(shù)f(x,y)=ln(x^2+y^2)，求f(x,y)在點(diǎn)(1,0)處的全微分。5.設(shè)函數(shù)f(x,y)=arctan(x/y)，求f(x,y)在點(diǎn)(1,1)處的偏導(dǎo)數(shù)。四、證明題要求：證明以下等式成立：證明：設(shè)函數(shù)f(x,y)=x^2y-y^2x，證明在點(diǎn)(0,0)處，f_x'(0,0)+f_y'(0,0)=0。五、計(jì)算題要求：計(jì)算下列偏導(dǎo)數(shù)和全微分：1.計(jì)算函數(shù)f(x,y)=e^(x^2+y^2)在點(diǎn)(1,0)處的偏導(dǎo)數(shù)f_x'(1,0)和f_y'(1,0)。2.計(jì)算函數(shù)f(x,y)=ln(x^2+y^2)在點(diǎn)(2,3)處的全微分df(2,3)。六、應(yīng)用題要求：求解下列問題：1.設(shè)函數(shù)f(x,y)=x^3+y^3，求在點(diǎn)(2,3)處的切平面方程。2.設(shè)函數(shù)f(x,y)=x^2y-y^2x，求在點(diǎn)(1,1)處的法線方程。本次試卷答案如下：一、選擇題1.B解析：f_x'(x,y)=2xy-y^2，代入x=1,y=1，得f_x'(1,1)=2*1*1-1^2=2-1=1。2.A解析：f_x'(x,y)=e^x*2y，代入x=0,y=1，得f_x'(0,1)=e^0*2*1=1*2=2。3.B解析：f_x'(x,y)=2x，f_y'(x,y)=2y，代入x=0,y=0，得df(0,0)=2*0*0+2*0*0=0。4.B解析：f_x'(x,y)=2x/(x^2+y^2)，f_y'(x,y)=-2y/(x^2+y^2)，代入x=1,y=0，得df(1,0)=2*1*0/(1^2+0^2)-2*0*0/(1^2+0^2)=0。5.D解析：f_x'(x,y)=(y/(x^2+y^2))'=-y^2/(x^2+y^2)^2，f_y'(x,y)=(x/(x^2+y^2))'=-x^2/(x^2+y^2)^2，代入x=1,y=1，得f_x'(1,1)=-1/(1+1)^2=-1/4，f_y'(1,1)=-1/(1+1)^2=-1/4，f_x'(1,1)+f_y'(1,1)=-1/4-1/4=-1/2，無定義。6.B解析：f_x'(x,y)=2xe^(x^2+y^2)，f_y'(x,y)=2ye^(x^2+y^2)，代入x=0,y=0，得df(0,0)=2*0*0+2*0*0=0。7.A解析：f_x'(x,y)=2xcos(x^2+y^2)，f_y'(x,y)=2ycos(x^2+y^2)，代入x=0,y=0，得df(0,0)=2*0*0+2*0*0=0。8.B解析：f_x'(x,y)=2x/(x^2-y^2)，f_y'(x,y)=-2y/(x^2-y^2)，代入x=1,y=0，得df(1,0)=2*1*0/(1^2-0^2)-2*0*0/(1^2-0^2)=0。9.C解析：f_x'(x,y)=-2x/(x^2+y^2)，f_y'(x,y)=-2y/(x^2+y^2)，代入x=1,y=0，得f_x'(1,0)=-2*1/(1^2+0^2)=-2。10.B解析：f_x'(x,y)=3x^2y^2，f_y'(x,y)=3x^2y^2，代入x=0,y=0，得df(0,0)=3*0*0*0+3*0*0*0=0。二、填空題1.df(1,1)=2dx+2dy解析：f_x'(x,y)=2xy，f_y'(x,y)=2xy，代入x=1,y=1，得df(1,1)=2*1*1dx+2*1*1dy=2dx+2dy。2.f_x'(0,0)=0解析：f_x'(x,y)=2xe^(x^2+y^2)，代入x=0,y=0，得f_x'(0,0)=2*0*0=0。3.f_y'(1,1)=-1解析：f_y'(x,y)=-x/(x^2+y^2)，代入x=1,y=1，得f_y'(1,1)=-1/(1^2+1^2)=-1/2。4.df(1,1)=2dx+2dy解析：f_x'(x,y)=2xy-y^2，f_y'(x,y)=x^2-2xy，代入x=1,y=1，得df(1,1)=(2*1*1-1^2)dx+(1^2-2*1*1)dy=2dx+2dy。5.df(1,0)=dx解析：f_x'(x,y)=2x/(x^2+y^2)，f_y'(x,y)=-2y/(x^2+y^2)，代入x=1,y=0，得df(1,0)=2*1*0/(1^2+0^2)dx+(-2*0*0)/(1^2+0^2)dy=dx。三、解答題1.df(2,3)=2dx+6dy解析：f_x'(x,y)=2x+2y，f_y'(x,y)=2x+2y，代入x=2,y=3，得df(2,3)=(2*2+2*3)dx+(2*2+2*3)dy=2dx+6dy。2.df(0,0)=0解析：f_x'(x,y)=2xe^(x^2+y^2)，f_y'(x,y)=2ye^(x^2+y^2)，代入x=0,y=0，得df(0,0)=2*0*0+2*0*0=0。3.df(1,1)=2dx+2dy解析：f_x'(x,y)=2xy-y^2，f_y'(x,y)=x^2-2xy，代入x=1,y=1，得df(1,1)=(2*1*1-1^2)dx+(1^2-2*1*1)dy=2dx+2dy。4.df(1,0)=dx解析：f_x'(x,y)=2x/(x^2+y^2)，f_y'(x,y)=-2y/(x^2+y^2)，代入x=1,y=0，得df(1,0)=2*1*0/(1^2+0^2)dx+(-2*0*0)/(1^2+0^2)dy=dx。5.f_x'(1,1)=-1/2，f_y'(1,1)=-1/2解析：f_x'(x,y)=(y/(x^2+y^2))'=-y^2/(x^2+y^2)^2，f_y'(x,y)=(x/(x^2+y^2))'=-x^2/(x^2+y^2)^2，代入x=1,y=1，得f_x'(1,1)=-1/(1+1)^2=-1/4，f_y'(1,1)=-1/(1+1)^2=-1/4。四、證明題證明：f_x'(x,y)=2xy-y^2，f_y'(x,y)=x^2-2xy，代入x=0,y=0，得f_x'(0,0)=0，f_y'(0,0)=0，因此f_x'(0,0)+f_y'(0,0)=0。五、計(jì)算題1.f_x'(1,0)=2，f_y'(1,0)=0解析：f_x'(x,y)=2xe^(x^2+y^2)，f_y'(x,y)=2ye^(x^2+y^2)，代入x=1,y=0，得f_x'(1,0)=2*1*0=0，f_y'(1,0)=2*0*0=0。2.df(2,3)=2dx+6dy解析：f_x'(x,y)=2x+2y，f_y'(x,y)=2x+2y，代入x=2,y=3，得df(2,3)=(2*2+2*3)dx+(2*2+2*3)dy=2dx+6dy。六、