三种方法计算形如函数y=√(ax+b).(cx^2+d)一阶导数过程步骤B16

函数y=√[(8x+1)/(25x2+15)]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=√[(8x+1)/(25x2+15)]的一阶导数的具体步骤。思路一：幂函数导数法∵y=√[(8x+1)/(25x2+15)]=[(8x+1)/(25x2+15)]^(1/2),∴y'=(1/2)*[(8x+1)/(25x2+15)]^(-1/2)*[8(25x2+15)-50x(8x+1)]/(25x2+15)2=(-1/2)*[(200x2+50x-120)/√[(8x+1)(25x2+15)]3。思路二：取对数求导数法∵y=√[(8x+1)/(25x2+15)]，∴lny=(1/2)ln[(8x+1)/ln(25x2+15)],即：2lny=ln(8x+1)-ln(25x2+15),两边同时对x求导得：2y'/y=8/(8x+1)-50x/(25x2+15)。进一步化简有y'=1/2·y·[8(25x2+15)-50x(8x+1)]/[(8x+1)(25x2+15)]=-1/2·y·(200x2+50x-120)/[(8x+1)(25x2+15)],=-1/2·√[(8x+1)/(25x2+15)]·(200x2+50x-120)/[(8x+1)(25x2+15)],=1/2·(200x2+50x-120)/[3√((8x+1)(25x2+15)]3.思路三：定义法求解导数∵f(x)=√[(8x+1)/(25x2+15)],∴y'=lim(t→0)[f(x+t)-f(x)]/t=lim(t→0)[√[(8x+8t+1)/(25x(x+t)2+15)-√[(8x+1)/(25x2+15)]/t=lim(t→0)[t(120-200x2-50x+200xt+25t)]/[2(25x2+15)2√[(8x+1)/(25x2+15)].y'=(120-200x2-50x+0)/{2(25x2+15)2√[(8x+1)/(25x2+15)]}=(-1/2)·(200x2+50x-120)/{√[(8x+1)(25x2+15)]3}。函数y=eq\r(\f(8x+1,25x2+15))一阶导数三种求法思路一：幂函数导数法y'=eq\f(1,2)·eq(\f(25x2+15,8x+1))\s\up12(\f(1,2))·eq\f(8(25x2+15)-50x(8x+1),(25x2+15)2),=eq\f(1,2)·eq(\f(25x2+15,8x+1))\s\up12(\f(1,2))·eq\f(120-200x2-50x,(25x2+15)2)=eq-\f(1,2)·eq\f(200x2+50x-120,\r((8x+1)(25x2+15)3))。思路二：取对数求导数法∵y=eq\r(\f(8x+1,25x2+15))，∴lny=eq\f(1,2)eqln\f(8x+1,25x2+15),即：2lny=ln(8x+1)-ln(25x2+15),两边同时对x求导得：eq\f(2y',y)=eq\f(8,8x+1)-eq\f(50x,25x2+15),则：y'=eq\f(1,2)·y·eq\f(8(25x2+15)-50x(8x+1),(8x+1)(25x2+15)),=-eq\f(1,2)·y·eq\f(200x2+50x-120,(8x+1)(25x2+15)),=-eq\f(1,2)·eq\r(\f(8x+1,25x2+15))·eq\f(200x2+50x-120,(8x+1)(25x2+15)),=-eq\f(1,2)·eq\f(200x2+50x-120,\r((8x+1)(25x2+15)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(8(x+t)+1,25(x+t)2+15))-eq\r(\f(8x+1,25x2+15)),t),=eq\a\ac(lim,t→0)eq\f(eq\f(8(x+t)+1,25(x+t)2+15)-eq\f(8x+1,25x2+15),t(eq\r(\f(8(x+t)+1,25(x+t)2+15))+eq\r(\f(8x+1,25x2+15)))),=eq\a\ac(lim,t→0)eq\f(\f([8(x+t)+1](25x2+15)-[25(x+t)2+15](8x+1),[25(x+t)2+15](25x2+15)),t(eq\r(\f(8(x+t)+1,25(x+t)2+15))+eq\r(\f(8x+1,25x2+15)))),=eq\a\ac(lim,t→0)eq\f(\f([8(x+t)+1](25x2+15)-[25(x+t)2+15](8x+1),(25x2+15)2),2teq\r(\f(8x+1,25x2+15))),=eq\a\ac(lim,t→0)eq\f(t(120-200x2-50x+200xt+25t),2(25x2+15)2eq\r(\f(8x+1,25x2+15)))=eq