三种方法计算形如函数y=√(ax+b).(cx^2+d)一阶导数过程步骤B5

函数y=√[(14x+19)/(2x2+25)]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=√[(14x+19)/(2x2+25)]的一阶导数的具体步骤。思路一：幂函数导数法∵y=√[(14x+19)/(2x2+25)]=[(14x+19)/(2x2+25)]^(1/2),∴y'=(1/2)*[(14x+19)/(2x2+25)]^(-1/2)*[14(2x2+25)-4x(14x+19)]/(2x2+25)2=(-1/2)*[(28x2+76x-350)/√[(14x+19)(2x2+25)]3。思路二：取对数求导数法∵y=√[(14x+19)/(2x2+25)]，∴lny=(1/2)ln[(14x+19)/ln(2x2+25)],即：2lny=ln(14x+19)-ln(2x2+25),两边同时对x求导得：2y'/y=14/(14x+19)-4x/(2x2+25)。进一步化简有y'=1/2·y·[14(2x2+25)-4x(14x+19)]/[(14x+19)(2x2+25)]=-1/2·y·(28x2+76x-350)/[(14x+19)(2x2+25)],=-1/2·√[(14x+19)/(2x2+25)]·(28x2+76x-350)/[(14x+19)(2x2+25)],=1/2·(28x2+76x-350)/[3√((14x+19)(2x2+25)]3.思路三：定义法求解导数∵f(x)=√[(14x+19)/(2x2+25)],∴y'=lim(t→0)[f(x+t)-f(x)]/t=lim(t→0)[√[(14x+8t+19)/(2x(x+t)2+25)-√[(14x+19)/(2x2+25)]/t=lim(t→0)[t(350-28x2-76x+28xt+38t)]/[2(2x2+25)2√[(14x+19)/(2x2+25)].y'=(350-28x2-76x+0)/{2(2x2+25)2√[(14x+19)/(2x2+25)]}=(-1/2)·(28x2+76x-350)/{√[(14x+19)(2x2+25)]3}。函数y=eq\r(\f(14x+19,2x2+25))一阶导数三种求法思路一：幂函数导数法y'=eq\f(1,2)·eq(\f(2x2+25,14x+19))\s\up12(\f(1,2))·eq\f(14(2x2+25)-4x(14x+19),(2x2+25)2),=eq\f(1,2)·eq(\f(2x2+25,14x+19))\s\up12(\f(1,2))·eq\f(350-28x2-76x,(2x2+25)2)=eq-\f(1,2)·eq\f(28x2+76x-350,\r((14x+19)(2x2+25)3))。思路二：取对数求导数法∵y=eq\r(\f(14x+19,2x2+25))，∴lny=eq\f(1,2)eqln\f(14x+19,2x2+25),即：2lny=ln(14x+19)-ln(2x2+25),两边同时对x求导得：eq\f(2y',y)=eq\f(14,14x+19)-eq\f(4x,2x2+25),则：y'=eq\f(1,2)·y·eq\f(14(2x2+25)-4x(14x+19),(14x+19)(2x2+25)),=-eq\f(1,2)·y·eq\f(28x2+76x-350,(14x+19)(2x2+25)),=-eq\f(1,2)·eq\r(\f(14x+19,2x2+25))·eq\f(28x2+76x-350,(14x+19)(2x2+25)),=-eq\f(1,2)·eq\f(28x2+76x-350,\r((14x+19)(2x2+25)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(14(x+t)+19,2(x+t)2+25))-eq\r(\f(14x+19,2x2+25)),t),=eq\a\ac(lim,t→0)eq\f(eq\f(14(x+t)+19,2(x+t)2+25)-eq\f(14x+19,2x2+25),t(eq\r(\f(14(x+t)+19,2(x+t)2+25))+eq\r(\f(14x+19,2x2+25)))),=eq\a\ac(lim,t→0)eq\f(\f([14(x+t)+19](2x2+25)-[2(x+t)2+25](14x+19),[2(x+t)2+25](2x2+25)),t(eq\r(\f(14(x+t)+19,2(x+t)2+25))+eq\r(\f(14x+19,2x2+25)))),=eq\a\ac(lim,t→0)eq\f(\f([14(x+t)+19](2x2+25)-[2(x+t)2+25](14x+19),(2x2+25)2),2teq\r(\f(14x+19,2x2+25))),=eq\a\ac(lim,t→0)eq\f(t(350-28x2-76x+28xt+38t),2(2x2+25)2eq\r(\f(14x+19,2x2+25)))=