三种方法计算形如函数y=√(ax+b).(cx^2+d)一阶导数过程步骤B12

函数y=√[(29x+13)/(28x2+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=√[(29x+13)/(28x2+30)]的一阶导数的具体步骤。思路一：幂函数导数法∵y=√[(29x+13)/(28x2+30)]=[(29x+13)/(28x2+30)]^(1/2),∴y'=(1/2)*[(29x+13)/(28x2+30)]^(-1/2)*[29(28x2+30)-56x(29x+13)]/(28x2+30)2=(-1/2)*[(812x2+728x-870)/√[(29x+13)(28x2+30)]3。思路二：取对数求导数法∵y=√[(29x+13)/(28x2+30)]，∴lny=(1/2)ln[(29x+13)/ln(28x2+30)],即：2lny=ln(29x+13)-ln(28x2+30),两边同时对x求导得：2y'/y=29/(29x+13)-56x/(28x2+30)。进一步化简有y'=1/2·y·[29(28x2+30)-56x(29x+13)]/[(29x+13)(28x2+30)]=-1/2·y·(812x2+728x-870)/[(29x+13)(28x2+30)],=-1/2·√[(29x+13)/(28x2+30)]·(812x2+728x-870)/[(29x+13)(28x2+30)],=1/2·(812x2+728x-870)/[3√((29x+13)(28x2+30)]3.思路三：定义法求解导数∵f(x)=√[(29x+13)/(28x2+30)],∴y'=lim(t→0)[f(x+t)-f(x)]/t=lim(t→0)[√[(29x+8t+13)/(28x(x+t)2+30)-√[(29x+13)/(28x2+30)]/t=lim(t→0)[t(870-812x2-728x+812xt+364t)]/[2(28x2+30)2√[(29x+13)/(28x2+30)].y'=(870-812x2-728x+0)/{2(28x2+30)2√[(29x+13)/(28x2+30)]}=(-1/2)·(812x2+728x-870)/{√[(29x+13)(28x2+30)]3}。函数y=eq\r(\f(29x+13,28x2+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(28x2+30,29x+13))\s\up12(\f(1,2))·eq\f(29(28x2+30)-56x(29x+13),(28x2+30)2),=eq\f(1,2)·eq(\f(28x2+30,29x+13))\s\up12(\f(1,2))·eq\f(870-812x2-728x,(28x2+30)2)=eq-\f(1,2)·eq\f(812x2+728x-870,\r((29x+13)(28x2+30)3))。思路二：取对数求导数法∵y=eq\r(\f(29x+13,28x2+30))，∴lny=eq\f(1,2)eqln\f(29x+13,28x2+30),即：2lny=ln(29x+13)-ln(28x2+30),两边同时对x求导得：eq\f(2y',y)=eq\f(29,29x+13)-eq\f(56x,28x2+30),则：y'=eq\f(1,2)·y·eq\f(29(28x2+30)-56x(29x+13),(29x+13)(28x2+30)),=-eq\f(1,2)·y·eq\f(812x2+728x-870,(29x+13)(28x2+30)),=-eq\f(1,2)·eq\r(\f(29x+13,28x2+30))·eq\f(812x2+728x-870,(29x+13)(28x2+30)),=-eq\f(1,2)·eq\f(812x2+728x-870,\r((29x+13)(28x2+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(29(x+t)+13,28(x+t)2+30))-eq\r(\f(29x+13,28x2+30)),t),=eq\a\ac(lim,t→0)eq\f(eq\f(29(x+t)+13,28(x+t)2+30)-eq\f(29x+13,28x2+30),t(eq\r(\f(29(x+t)+13,28(x+t)2+30))+eq\r(\f(29x+13,28x2+30)))),=eq\a\ac(lim,t→0)eq\f(\f([29(x+t)+13](28x2+30)-[28(x+t)2+30](29x+13),[28(x+t)2+30](28x2+30)),t(eq\r(\f(29(x+t)+13,28(x+t)2+30))+eq\r(\f(29x+13,28x2+30)))),=eq\a\ac(lim,t→0)eq\f(\f([29(x+t)+13](28x2+30)-[28(x+t)2+30](29x+13),(28x2+30)2),2teq\r(\f(29x+13,28x2+30))),=eq\a\ac(lim,t→0)eq\f(t(870-812x2-728x+812xt+364t),2(28x2+30)2eq\r(\f(29x+13,28x2+30)))=eq