三种方法计算形如函数y=√(ax+b).(cx^2+d)一阶导数过程步骤B11

函数y=√[(4x+11)/(23x2+5)]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=√[(4x+11)/(23x2+5)]的一阶导数的具体步骤。思路一：幂函数导数法∵y=√[(4x+11)/(23x2+5)]=[(4x+11)/(23x2+5)]^(1/2),∴y'=(1/2)*[(4x+11)/(23x2+5)]^(-1/2)*[4(23x2+5)-46x(4x+11)]/(23x2+5)2=(-1/2)*[(92x2+506x-20)/√[(4x+11)(23x2+5)]3。思路二：取对数求导数法∵y=√[(4x+11)/(23x2+5)]，∴lny=(1/2)ln[(4x+11)/ln(23x2+5)],即：2lny=ln(4x+11)-ln(23x2+5),两边同时对x求导得：2y'/y=4/(4x+11)-46x/(23x2+5)。进一步化简有y'=1/2·y·[4(23x2+5)-46x(4x+11)]/[(4x+11)(23x2+5)]=-1/2·y·(92x2+506x-20)/[(4x+11)(23x2+5)],=-1/2·√[(4x+11)/(23x2+5)]·(92x2+506x-20)/[(4x+11)(23x2+5)],=1/2·(92x2+506x-20)/[3√((4x+11)(23x2+5)]3.思路三：定义法求解导数∵f(x)=√[(4x+11)/(23x2+5)],∴y'=lim(t→0)[f(x+t)-f(x)]/t=lim(t→0)[√[(4x+8t+11)/(23x(x+t)2+5)-√[(4x+11)/(23x2+5)]/t=lim(t→0)[t(20-92x2-506x+92xt+253t)]/[2(23x2+5)2√[(4x+11)/(23x2+5)].y'=(20-92x2-506x+0)/{2(23x2+5)2√[(4x+11)/(23x2+5)]}=(-1/2)·(92x2+506x-20)/{√[(4x+11)(23x2+5)]3}。函数y=eq\r(\f(4x+11,23x2+5))一阶导数三种求法思路一：幂函数导数法y'=eq\f(1,2)·eq(\f(23x2+5,4x+11))\s\up12(\f(1,2))·eq\f(4(23x2+5)-46x(4x+11),(23x2+5)2),=eq\f(1,2)·eq(\f(23x2+5,4x+11))\s\up12(\f(1,2))·eq\f(20-92x2-506x,(23x2+5)2)=eq-\f(1,2)·eq\f(92x2+506x-20,\r((4x+11)(23x2+5)3))。思路二：取对数求导数法∵y=eq\r(\f(4x+11,23x2+5))，∴lny=eq\f(1,2)eqln\f(4x+11,23x2+5),即：2lny=ln(4x+11)-ln(23x2+5),两边同时对x求导得：eq\f(2y',y)=eq\f(4,4x+11)-eq\f(46x,23x2+5),则：y'=eq\f(1,2)·y·eq\f(4(23x2+5)-46x(4x+11),(4x+11)(23x2+5)),=-eq\f(1,2)·y·eq\f(92x2+506x-20,(4x+11)(23x2+5)),=-eq\f(1,2)·eq\r(\f(4x+11,23x2+5))·eq\f(92x2+506x-20,(4x+11)(23x2+5)),=-eq\f(1,2)·eq\f(92x2+506x-20,\r((4x+11)(23x2+5)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(4(x+t)+11,23(x+t)2+5))-eq\r(\f(4x+11,23x2+5)),t),=eq\a\ac(lim,t→0)eq\f(eq\f(4(x+t)+11,23(x+t)2+5)-eq\f(4x+11,23x2+5),t(eq\r(\f(4(x+t)+11,23(x+t)2+5))+eq\r(\f(4x+11,23x2+5)))),=eq\a\ac(lim,t→0)eq\f(\f([4(x+t)+11](23x2+5)-[23(x+t)2+5](4x+11),[23(x+t)2+5](23x2+5)),t(eq\r(\f(4(x+t)+11,23(x+t)2+5))+eq\r(\f(4x+11,23x2+5)))),=eq\a\ac(lim,t→0)eq\f(\f([4(x+t)+11](23x2+5)-[23(x+t)2+5](4x+11),(23x2+5)2),2teq\r(\f(4x+11,23x2+5))),=eq\a\ac(lim,t→0)eq\f(t(20-92x2-506x+92xt+253t),2(23x2+5)2eq\r(\f(4x+11,23x2+5)))=e