函数y(110x²+64x+19)=118的多阶导数计算过程

y=eq\f(118,110x2+64x+19)的一阶和二阶导数主要内容：本文通过函数商的求导和函数乘积的求导法则，介绍求函数y=eq\f(118,110x2+64x+19)一阶及二阶导数的主要过程和步骤。一、两种方法求一阶导数函数商求导法：y=eq\f(118,110x2+64x+19)y´=eq\f(118´(110x2+64x+19)-118(110x2+64x+19)´,(110x2+64x+19)2),=eq\f(0-25960x-7552,(110x2+64x+19)2),=-472*eq\f(55x+16,(110x2+64x+19)2).函数乘积求导法：y=eq\f(118,110x2+64x+19)，即：y(110x2+64x+19)=a,两边同时对x求导得：y´(110x2+64x+19)+y(220x+64)=0,y´(110x2+64x+19)=-y(220x+64)y´=-yeq\f(220x+64,110x2+64x+19),则：y´=-472*eq\f(55x+16,(110x2+64x+19)2).函数导数定义法：y=eq\f(118,110x2+64x+19)，则：y´=lim(t→0)eq\f(eq\f(118,110(x+t)2+64(x+t)+19)-eq\f(118,110x2+64x+19),t),=lim(t→0)eq\f(110x2+64x+19-110(x+t)2-64(x+t)-19,(110x2+64x+19)[110(x+t)2+64(x+t)+19]t),=lim(t→0)eq\f(-110(2xt+t2)-64t,(110x2+64x+19)[110(x+t)2+64(x+t)+19]t)=lim(t→0)eq\f(-110t(2x+t)-64,(110x2+64x+19)[110(x+t)2+64(x+t)+19]t),=lim(t→0)eq\f(-110(2x+t)-64,(110x2+64x+19)[110(x+t)2+64(x+t)+19]),=eq\f(-110(2x+0)-64,(110x2+64x+19)[110(x+0)2+64(x+0)+19]),=-472*eq\f(55x+16,(110x2+64x+19)2)。二、函数商求二阶导数直接求导法：∵y´=-472eq\f(55x+16,(110x2+64x+19)2)∴y´´=-472*eq\f(55(110x2+64x+19)-2(55x+16)(220x+64),(110x2+64x+19)3),=472*eq\f(2(55x+16)(220x+64)-55(110x2+64x+19),(110x2+64x+19)3),=472*eq\f(18150x2+10560x+1003,(110x2+64x+19)3)。函数乘积求导法：∵y´=-472eq\f(55x+16,(110x2+64x+19)2),即：y´(110x2+64x+19)2=-472(55x+16)∴y´´(110x2+64x+19)²+y´*4*(110x2+64x+19)*110x=-472*55，y´´(110x2+64x+19)²=-472*55-y´*4*(110x2+64x+19)*110x，将y´=-472eq\f(55x+16,(110x2+64x+19)2)代入上式，有：y´´(110x2+64x+19)²=-472*55+472eq\f(55x+16,(110x2+64x+19)2)*4*(110x2+64x+19)*110x，y´´(110x2+64x+19)²=-472*55+472eq\f(55x+16,110x2+64x+19)*4*110x，化简得：y´´=472*eq\f(18150x2+10560x+1003,(110x2+64x+19)3)。导数定义法：∵y´=-472eq\f(55x+16,(110x2+64x+19)2)∴y´´=-472EQ\s(lim,t→0)eq\f(eq\f(55(x+t)+16,[110(x+t)2+64(x+t)+19]2)-eq\f(55x+16,(110x2+64x+19)2),t),=-472EQ\s(lim,t→0)eq\f([55(x+t)+16](110x2+64x+19)2-(55x+16)[110(x+t)2+64(x+t)+19]2,t(110x2+64x+19)2[110(x+t)2+64(x+t)+19]2),=472EQ\s(lim,t→0)eq\f((55x+16){[110(x+t)2+64(x+t)+19]2-(110x2+64x+19)2}-55t(110x2+64x+19)2,t(110x2+64x+19)2[110(x+t)2+64(x+t)+19]2),=472EQ\s(lim,t→0)eq\f((55x+16)t(110t+2*110x+64)[110(x+t)2+64(x+t)+19+110x2+64x+19]-55t(110x2+64x+19)2,t(110x2+64x+19)2[110(x+t)2+64(x+t)+19]2)=472EQ\s(lim,t→0)eq\f((55x+16)(110t+2*110x+64)[110(x+t)2+64(x+t)+19+110x2+64x+19]-55(110x2+64x+19)2,(110x2+64x+19)2[110(x+t)2+64(x+t)+19]2)=472*eq\f(2*(55x+16)(2*110x+64)(110x2+64x+19)-55(110x2+64x+19)2,(110x2+64x+19)2(110x2+64x+19)2)