三种方法计算形如函数y=√(ax+b).(cx^2+d)一阶导数过程步骤B3

函数y=√[(6x+11)/(9x2+30)]一阶导数三种求法.主要内容：本文通过幂函数导数法、取对数法和导数定义法，介绍求解函数y=√[(6x+11)/(9x2+30)]的一阶导数的具体步骤。思路一：幂函数导数法∵y=√[(6x+11)/(9x2+30)]=[(6x+11)/(9x2+30)]^(1/2),∴y'=(1/2)*[(6x+11)/(9x2+30)]^(-1/2)*[6(9x2+30)-18x(6x+11)]/(9x2+30)2=(-1/2)*[(54x2+198x-180)/√[(6x+11)(9x2+30)]3。思路二：取对数求导数法∵y=√[(6x+11)/(9x2+30)]，∴lny=(1/2)ln[(6x+11)/ln(9x2+30)],即：2lny=ln(6x+11)-ln(9x2+30),两边同时对x求导得：2y'/y=6/(6x+11)-18x/(9x2+30)。进一步化简有y'=1/2·y·[6(9x2+30)-18x(6x+11)]/[(6x+11)(9x2+30)]=-1/2·y·(54x2+198x-180)/[(6x+11)(9x2+30)],=-1/2·√[(6x+11)/(9x2+30)]·(54x2+198x-180)/[(6x+11)(9x2+30)],=1/2·(54x2+198x-180)/[3√((6x+11)(9x2+30)]3.思路三：定义法求解导数∵f(x)=√[(6x+11)/(9x2+30)],∴y'=lim(t→0)[f(x+t)-f(x)]/t=lim(t→0)[√[(6x+8t+11)/(9x(x+t)2+30)-√[(6x+11)/(9x2+30)]/t=lim(t→0)[t(180-54x2-198x+54xt+99t)]/[2(9x2+30)2√[(6x+11)/(9x2+30)].y'=(180-54x2-198x+0)/{2(9x2+30)2√[(6x+11)/(9x2+30)]}=(-1/2)·(54x2+198x-180)/{√[(6x+11)(9x2+30)]3}。函数y=eq\r(\f(6x+11,9x2+30))一阶导数三种求法思路一：幂函数导数法y'=eq\f(1,2)·eq(\f(9x2+30,6x+11))\s\up12(\f(1,2))·eq\f(6(9x2+30)-18x(6x+11),(9x2+30)2),=eq\f(1,2)·eq(\f(9x2+30,6x+11))\s\up12(\f(1,2))·eq\f(180-54x2-198x,(9x2+30)2)=eq-\f(1,2)·eq\f(54x2+198x-180,\r((6x+11)(9x2+30)3))。思路二：取对数求导数法∵y=eq\r(\f(6x+11,9x2+30))，∴lny=eq\f(1,2)eqln\f(6x+11,9x2+30),即：2lny=ln(6x+11)-ln(9x2+30),两边同时对x求导得：eq\f(2y',y)=eq\f(6,6x+11)-eq\f(18x,9x2+30),则：y'=eq\f(1,2)·y·eq\f(6(9x2+30)-18x(6x+11),(6x+11)(9x2+30)),=-eq\f(1,2)·y·eq\f(54x2+198x-180,(6x+11)(9x2+30)),=-eq\f(1,2)·eq\r(\f(6x+11,9x2+30))·eq\f(54x2+198x-180,(6x+11)(9x2+30)),=-eq\f(1,2)·eq\f(54x2+198x-180,\r((6x+11)(9x2+30)3))。思路三：定义法求解导数y'=eq\a\ac(lim,t→0)eq\f(f(x+t)-f(x),t)=eq\a\ac(lim,t→0)eq\f(eq\r(\f(6(x+t)+11,9(x+t)2+30))-eq\r(\f(6x+11,9x2+30)),t),=eq\a\ac(lim,t→0)eq\f(eq\f(6(x+t)+11,9(x+t)2+30)-eq\f(6x+11,9x2+30),t(eq\r(\f(6(x+t)+11,9(x+t)2+30))+eq\r(\f(6x+11,9x2+30)))),=eq\a\ac(lim,t→0)eq\f(\f([6(x+t)+11](9x2+30)-[9(x+t)2+30](6x+11),[9(x+t)2+30](9x2+30)),t(eq\r(\f(6(x+t)+11,9(x+t)2+30))+eq\r(\f(6x+11,9x2+30)))),=eq\a\ac(lim,t→0)eq\f(\f([6(x+t)+11](9x2+30)-[9(x+t)2+30](6x+11),(9x2+30)2),2teq\r(\f(6x+11,9x2+30))),=eq\a\ac(lim,t→0)eq\f(t(180-54x2-198x+54xt+99t),2(9x2+30)2eq\r(\f(6x+11,9x2+30)))=eq