函數(shù)商類型二階偏導(dǎo)數(shù)計(jì)算練習(xí)及步驟?1：求y=eq\f(x-49,x+16)的二階導(dǎo)數(shù)。?2：函數(shù)y=eq\f(3,x2-32x+75)的二階導(dǎo)數(shù)。?3：求y=eq\f(18x,33+x2)的二階導(dǎo)數(shù)。?4：計(jì)算y=eq\f(sin8x,x+10)的二階導(dǎo)數(shù)。?5：求y=4x+eq\f(x,x2-31)的二階導(dǎo)數(shù)。函數(shù)商類型二階偏導(dǎo)數(shù)計(jì)算練習(xí)及步驟參考?1：求y=eq\f(x-49,x+16)的二階導(dǎo)數(shù)。解：法一：y=eq\f(x-49,x+16),y'=eq\f(x+16-x+49,(x+16)2)=eq\f(65,(x+16)2);y"=-eq\f(130,(x+16)3)。法二：方程變形為y(x+16)=x-49,兩邊求導(dǎo)有：y'(x+16)+y=1,再次求導(dǎo)：y"(x+16)+y'+y'=0,y"(x+16)=-2y'=-2*eq\f(1-y,x+16),y"=-2*eq\f(1-y,(x+16)2)=-eq\f(130,(x+16)3)。?2：函數(shù)y=eq\f(3,x2-32x+75)的二階導(dǎo)數(shù)。解：y=eq\f(3,x2-32x+75)，用商求導(dǎo)法則有：y'=-3*eq\f(2x-32,(x2-32x+75)2);y"=-3*eq\f(2(x2-32x+75)2-2(2x-32)2(x2-32x+75),(x2-32x+75)4),=-6*eq\f((x2-32x+75)-(2x-32)2,(x2-32x+75)3),=6*eq\f(3x2+96x+36,(x2-32x+75)3)。?3：求y=eq\f(18x,33+x2)的二階導(dǎo)數(shù)。解：法一：y=eq\f(18x,33+x2)，使用商求導(dǎo)法則：y'=eq\f(18(33+x2)-18x*2x,(33+x2)2)=eq\f(18(33-x2),(33+x2)2);y"=eq\f(18[(-2x)(33+x2)-2*(33-x2)*2x],(33+x2)3),=-36*eq\f(x(33+x2)+2(33-x2)x,(33+x2)3),=36*eq\f(x(x2-3*33),(33+x2)3)。法二：乘積法，此時(shí)y(33+x2)=18x,兩邊求導(dǎo)有：y'(33+x2)+2xy=18,再次求導(dǎo)有：y"(33+x2)+2xy'+2y+2xy'=0,y"=-eq\f(2(2xy'+y),(33+x2)),再將y和y'代入化簡，y"=36*eq\f(x(x2-3*33),(33+x2)3)。?4：計(jì)算y=eq\f(sin8x,x+10)的二階導(dǎo)數(shù)。解：y=eq\f(sin8x,x+10)，y'=eq\f(8cos8x(x+10)-sin8x,(x+10)2)=eq\f(A,B);y"=eq\f(A'B-AB',B2),其中：A'=-82sin8x(x+10)+8cos8x-8cos8x=-82sin8x(x+10),B'=2(x+10),y"=eq\f(-82sin8x(x+10)(x+10)2-2[8cos8x(x+10)-sin8x](x+10),(x+10)4),=eq\f(-82sin8x(x+10)(x+10)-2[8cos8x(x+10)-sin8x],(x+10)3)?5：求y=4x+eq\f(x,x2-31)的二階導(dǎo)數(shù)。解：y=4x+eq\f(x,x2-31)，y'=4+eq\f(x2-31-2x2,(x2-31)2)=4-eq\f(31+x2,(x2-3