三种方法计算形如函数y=√(ax+b).(cx^2+d)一阶导数过程步骤B8

函数y=√[(29x+10)/(14x2+10)]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=√[(29x+10)/(14x2+10)]的一阶导数的具体步骤。思路一：幂函数导数法∵y=√[(29x+10)/(14x2+10)]=[(29x+10)/(14x2+10)]^(1/2),∴y'=(1/2)*[(29x+10)/(14x2+10)]^(-1/2)*[29(14x2+10)-28x(29x+10)]/(14x2+10)2=(-1/2)*[(406x2+280x-290)/√[(29x+10)(14x2+10)]3。思路二：取对数求导数法∵y=√[(29x+10)/(14x2+10)]，∴lny=(1/2)ln[(29x+10)/ln(14x2+10)],即：2lny=ln(29x+10)-ln(14x2+10),两边同时对x求导得：2y'/y=29/(29x+10)-28x/(14x2+10)。进一步化简有y'=1/2·y·[29(14x2+10)-28x(29x+10)]/[(29x+10)(14x2+10)]=-1/2·y·(406x2+280x-290)/[(29x+10)(14x2+10)],=-1/2·√[(29x+10)/(14x2+10)]·(406x2+280x-290)/[(29x+10)(14x2+10)],=1/2·(406x2+280x-290)/[3√((29x+10)(14x2+10)]3.思路三：定义法求解导数∵f(x)=√[(29x+10)/(14x2+10)],∴y'=lim(t→0)[f(x+t)-f(x)]/t=lim(t→0)[√[(29x+8t+10)/(14x(x+t)2+10)-√[(29x+10)/(14x2+10)]/t=lim(t→0)[t(290-406x2-280x+406xt+140t)]/[2(14x2+10)2√[(29x+10)/(14x2+10)].y'=(290-406x2-280x+0)/{2(14x2+10)2√[(29x+10)/(14x2+10)]}=(-1/2)·(406x2+280x-290)/{√[(29x+10)(14x2+10)]3}。函数y=eq\r(\f(29x+10,14x2+10))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+10,29x+10))\s\up12(\f(1,2))·eq\f(29(14x2+10)-28x(29x+10),(14x2+10)2),=eq\f(1,2)·eq(\f(14x2+10,29x+10))\s\up12(\f(1,2))·eq\f(290-406x2-280x,(14x2+10)2)=eq-\f(1,2)·eq\f(406x2+280x-290,\r((29x+10)(14x2+10)3))。思路二：取对数求导数法∵y=eq\r(\f(29x+10,14x2+10))，∴lny=eq\f(1,2)eqln\f(29x+10,14x2+10),即：2lny=ln(29x+10)-ln(14x2+10),两边同时对x求导得：eq\f(2y',y)=eq\f(29,29x+10)-eq\f(28x,14x2+10),则：y'=eq\f(1,2)·y·eq\f(29(14x2+10)-28x(29x+10),(29x+10)(14x2+10)),=-eq\f(1,2)·y·eq\f(406x2+280x-290,(29x+10)(14x2+10)),=-eq\f(1,2)·eq\r(\f(29x+10,14x2+10))·eq\f(406x2+280x-290,(29x+10)(14x2+10)),=-eq\f(1,2)·eq\f(406x2+280x-290,\r((29x+10)(14x2+10)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(29(x+t)+10,14(x+t)2+10))-eq\r(\f(29x+10,14x2+10)),t),=eq\a\ac(lim,t→0)eq\f(eq\f(29(x+t)+10,14(x+t)2+10)-eq\f(29x+10,14x2+10),t(eq\r(\f(29(x+t)+10,14(x+t)2+10))+eq\r(\f(29x+10,14x2+10)))),=eq\a\ac(lim,t→0)eq\f(\f([29(x+t)+10](14x2+10)-[14(x+t)2+10](29x+10),[14(x+t)2+10](14x2+10)),t(eq\r(\f(29(x+t)+10,14(x+t)2+10))+eq\r(\f(29x+10,14x2+10)))),=eq\a\ac(lim,t→0)eq\f(\f([29(x+t)+10](14x2+10)-[14(x+t)2+10](29x+10),(14x2+10)2),2teq\r(\f(29x+10,14x2+10))),=eq\a\ac(lim,t→0)eq\f(t(290-406x2-280x+406xt+140t),2(14x2+10)2eq\r(\f(29x+10,14x2+10)))=eq