三种方法计算形如函数y=√(ax+b).(cx^2+d)一阶导数过程步骤B18

函数y=√[(13x+15)/(7x2+27)]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=√[(13x+15)/(7x2+27)]的一阶导数的具体步骤。思路一：幂函数导数法∵y=√[(13x+15)/(7x2+27)]=[(13x+15)/(7x2+27)]^(1/2),∴y'=(1/2)*[(13x+15)/(7x2+27)]^(-1/2)*[13(7x2+27)-14x(13x+15)]/(7x2+27)2=(-1/2)*[(91x2+210x-351)/√[(13x+15)(7x2+27)]3。思路二：取对数求导数法∵y=√[(13x+15)/(7x2+27)]，∴lny=(1/2)ln[(13x+15)/ln(7x2+27)],即：2lny=ln(13x+15)-ln(7x2+27),两边同时对x求导得：2y'/y=13/(13x+15)-14x/(7x2+27)。进一步化简有y'=1/2·y·[13(7x2+27)-14x(13x+15)]/[(13x+15)(7x2+27)]=-1/2·y·(91x2+210x-351)/[(13x+15)(7x2+27)],=-1/2·√[(13x+15)/(7x2+27)]·(91x2+210x-351)/[(13x+15)(7x2+27)],=1/2·(91x2+210x-351)/[3√((13x+15)(7x2+27)]3.思路三：定义法求解导数∵f(x)=√[(13x+15)/(7x2+27)],∴y'=lim(t→0)[f(x+t)-f(x)]/t=lim(t→0)[√[(13x+8t+15)/(7x(x+t)2+27)-√[(13x+15)/(7x2+27)]/t=lim(t→0)[t(351-91x2-210x+91xt+105t)]/[2(7x2+27)2√[(13x+15)/(7x2+27)].y'=(351-91x2-210x+0)/{2(7x2+27)2√[(13x+15)/(7x2+27)]}=(-1/2)·(91x2+210x-351)/{√[(13x+15)(7x2+27)]3}。函数y=eq\r(\f(13x+15,7x2+27))一阶导数三种求法思路一：幂函数导数法y'=eq\f(1,2)·eq(\f(7x2+27,13x+15))\s\up12(\f(1,2))·eq\f(13(7x2+27)-14x(13x+15),(7x2+27)2),=eq\f(1,2)·eq(\f(7x2+27,13x+15))\s\up12(\f(1,2))·eq\f(351-91x2-210x,(7x2+27)2)=eq-\f(1,2)·eq\f(91x2+210x-351,\r((13x+15)(7x2+27)3))。思路二：取对数求导数法∵y=eq\r(\f(13x+15,7x2+27))，∴lny=eq\f(1,2)eqln\f(13x+15,7x2+27),即：2lny=ln(13x+15)-ln(7x2+27),两边同时对x求导得：eq\f(2y',y)=eq\f(13,13x+15)-eq\f(14x,7x2+27),则：y'=eq\f(1,2)·y·eq\f(13(7x2+27)-14x(13x+15),(13x+15)(7x2+27)),=-eq\f(1,2)·y·eq\f(91x2+210x-351,(13x+15)(7x2+27)),=-eq\f(1,2)·eq\r(\f(13x+15,7x2+27))·eq\f(91x2+210x-351,(13x+15)(7x2+27)),=-eq\f(1,2)·eq\f(91x2+210x-351,\r((13x+15)(7x2+27)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(13(x+t)+15,7(x+t)2+27))-eq\r(\f(13x+15,7x2+27)),t),=eq\a\ac(lim,t→0)eq\f(eq\f(13(x+t)+15,7(x+t)2+27)-eq\f(13x+15,7x2+27),t(eq\r(\f(13(x+t)+15,7(x+t)2+27))+eq\r(\f(13x+15,7x2+27)))),=eq\a\ac(lim,t→0)eq\f(\f([13(x+t)+15](7x2+27)-[7(x+t)2+27](13x+15),[7(x+t)2+27](7x2+27)),t(eq\r(\f(13(x+t)+15,7(x+t)2+27))+eq\r(\f(13x+15,7x2+27)))),=eq\a\ac(lim,t→0)eq\f(\f([13(x+t)+15](7x2+27)-[7(x+t)2+27](13x+15),(7x2+27)2),2teq\r(\f(13x+15,7x2+27))),=eq\a\ac(lim,t→0)eq\f(t(351-91x2-210x+91xt+105t),2(7x2+27)2eq\r(\f(13x+15,7x2+27)))=e