三种方法计算形如函数y=√(ax+b).(cx^2+d)一阶导数过程步骤B10

函数y=√[(19x+30)/(14x2+9)]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=√[(19x+30)/(14x2+9)]的一阶导数的具体步骤。思路一：幂函数导数法∵y=√[(19x+30)/(14x2+9)]=[(19x+30)/(14x2+9)]^(1/2),∴y'=(1/2)*[(19x+30)/(14x2+9)]^(-1/2)*[19(14x2+9)-28x(19x+30)]/(14x2+9)2=(-1/2)*[(266x2+840x-171)/√[(19x+30)(14x2+9)]3。思路二：取对数求导数法∵y=√[(19x+30)/(14x2+9)]，∴lny=(1/2)ln[(19x+30)/ln(14x2+9)],即：2lny=ln(19x+30)-ln(14x2+9),两边同时对x求导得：2y'/y=19/(19x+30)-28x/(14x2+9)。进一步化简有y'=1/2·y·[19(14x2+9)-28x(19x+30)]/[(19x+30)(14x2+9)]=-1/2·y·(266x2+840x-171)/[(19x+30)(14x2+9)],=-1/2·√[(19x+30)/(14x2+9)]·(266x2+840x-171)/[(19x+30)(14x2+9)],=1/2·(266x2+840x-171)/[3√((19x+30)(14x2+9)]3.思路三：定义法求解导数∵f(x)=√[(19x+30)/(14x2+9)],∴y'=lim(t→0)[f(x+t)-f(x)]/t=lim(t→0)[√[(19x+8t+30)/(14x(x+t)2+9)-√[(19x+30)/(14x2+9)]/t=lim(t→0)[t(171-266x2-840x+266xt+420t)]/[2(14x2+9)2√[(19x+30)/(14x2+9)].y'=(171-266x2-840x+0)/{2(14x2+9)2√[(19x+30)/(14x2+9)]}=(-1/2)·(266x2+840x-171)/{√[(19x+30)(14x2+9)]3}。函数y=eq\r(\f(19x+30,14x2+9))一阶导数三种求法思路一：幂函数导数法y'=eq\f(1,2)·eq(\f(14x2+9,19x+30))\s\up12(\f(1,2))·eq\f(19(14x2+9)-28x(19x+30),(14x2+9)2),=eq\f(1,2)·eq(\f(14x2+9,19x+30))\s\up12(\f(1,2))·eq\f(171-266x2-840x,(14x2+9)2)=eq-\f(1,2)·eq\f(266x2+840x-171,\r((19x+30)(14x2+9)3))。思路二：取对数求导数法∵y=eq\r(\f(19x+30,14x2+9))，∴lny=eq\f(1,2)eqln\f(19x+30,14x2+9),即：2lny=ln(19x+30)-ln(14x2+9),两边同时对x求导得：eq\f(2y',y)=eq\f(19,19x+30)-eq\f(28x,14x2+9),则：y'=eq\f(1,2)·y·eq\f(19(14x2+9)-28x(19x+30),(19x+30)(14x2+9)),=-eq\f(1,2)·y·eq\f(266x2+840x-171,(19x+30)(14x2+9)),=-eq\f(1,2)·eq\r(\f(19x+30,14x2+9))·eq\f(266x2+840x-171,(19x+30)(14x2+9)),=-eq\f(1,2)·eq\f(266x2+840x-171,\r((19x+30)(14x2+9)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(19(x+t)+30,14(x+t)2+9))-eq\r(\f(19x+30,14x2+9)),t),=eq\a\ac(lim,t→0)eq\f(eq\f(19(x+t)+30,14(x+t)2+9)-eq\f(19x+30,14x2+9),t(eq\r(\f(19(x+t)+30,14(x+t)2+9))+eq\r(\f(19x+30,14x2+9)))),=eq\a\ac(lim,t→0)eq\f(\f([19(x+t)+30](14x2+9)-[14(x+t)2+9](19x+30),[14(x+t)2+9](14x2+9)),t(eq\r(\f(19(x+t)+30,14(x+t)2+9))+eq\r(\f(19x+30,14x2+9)))),=eq\a\ac(lim,t→0)eq\f(\f([19(x+t)+30](14x2+9)-[14(x+t)2+9](19x+30),(14x2+9)2),2teq\r(\f(19x+30,14x2+9))),=eq\a\ac(lim,t→0)eq\f(t(171-266x2-840x+266xt+420t),2(14x2+9)2eq\r(\f(19x+30,14x2+9)))=eq