三种方法计算形如函数y=√(ax+b).(cx^2+d)一阶导数过程步骤B15

函数y=√[(17x+28)/(18x2+6)]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=√[(17x+28)/(18x2+6)]的一阶导数的具体步骤。思路一：幂函数导数法∵y=√[(17x+28)/(18x2+6)]=[(17x+28)/(18x2+6)]^(1/2),∴y'=(1/2)*[(17x+28)/(18x2+6)]^(-1/2)*[17(18x2+6)-36x(17x+28)]/(18x2+6)2=(-1/2)*[(306x2+1008x-102)/√[(17x+28)(18x2+6)]3。思路二：取对数求导数法∵y=√[(17x+28)/(18x2+6)]，∴lny=(1/2)ln[(17x+28)/ln(18x2+6)],即：2lny=ln(17x+28)-ln(18x2+6),两边同时对x求导得：2y'/y=17/(17x+28)-36x/(18x2+6)。进一步化简有y'=1/2·y·[17(18x2+6)-36x(17x+28)]/[(17x+28)(18x2+6)]=-1/2·y·(306x2+1008x-102)/[(17x+28)(18x2+6)],=-1/2·√[(17x+28)/(18x2+6)]·(306x2+1008x-102)/[(17x+28)(18x2+6)],=1/2·(306x2+1008x-102)/[3√((17x+28)(18x2+6)]3.思路三：定义法求解导数∵f(x)=√[(17x+28)/(18x2+6)],∴y'=lim(t→0)[f(x+t)-f(x)]/t=lim(t→0)[√[(17x+8t+28)/(18x(x+t)2+6)-√[(17x+28)/(18x2+6)]/t=lim(t→0)[t(102-306x2-1008x+306xt+504t)]/[2(18x2+6)2√[(17x+28)/(18x2+6)].y'=(102-306x2-1008x+0)/{2(18x2+6)2√[(17x+28)/(18x2+6)]}=(-1/2)·(306x2+1008x-102)/{√[(17x+28)(18x2+6)]3}。函数y=eq\r(\f(17x+28,18x2+6))一阶导数三种求法思路一：幂函数导数法y'=eq\f(1,2)·eq(\f(18x2+6,17x+28))\s\up12(\f(1,2))·eq\f(17(18x2+6)-36x(17x+28),(18x2+6)2),=eq\f(1,2)·eq(\f(18x2+6,17x+28))\s\up12(\f(1,2))·eq\f(102-306x2-1008x,(18x2+6)2)=eq-\f(1,2)·eq\f(306x2+1008x-102,\r((17x+28)(18x2+6)3))。思路二：取对数求导数法∵y=eq\r(\f(17x+28,18x2+6))，∴lny=eq\f(1,2)eqln\f(17x+28,18x2+6),即：2lny=ln(17x+28)-ln(18x2+6),两边同时对x求导得：eq\f(2y',y)=eq\f(17,17x+28)-eq\f(36x,18x2+6),则：y'=eq\f(1,2)·y·eq\f(17(18x2+6)-36x(17x+28),(17x+28)(18x2+6)),=-eq\f(1,2)·y·eq\f(306x2+1008x-102,(17x+28)(18x2+6)),=-eq\f(1,2)·eq\r(\f(17x+28,18x2+6))·eq\f(306x2+1008x-102,(17x+28)(18x2+6)),=-eq\f(1,2)·eq\f(306x2+1008x-102,\r((17x+28)(18x2+6)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(17(x+t)+28,18(x+t)2+6))-eq\r(\f(17x+28,18x2+6)),t),=eq\a\ac(lim,t→0)eq\f(eq\f(17(x+t)+28,18(x+t)2+6)-eq\f(17x+28,18x2+6),t(eq\r(\f(17(x+t)+28,18(x+t)2+6))+eq\r(\f(17x+28,18x2+6)))),=eq\a\ac(lim,t→0)eq\f(\f([17(x+t)+28](18x2+6)-[18(x+t)2+6](17x+28),[18(x+t)2+6](18x2+6)),t(eq\r(\f(17(x+t)+28,18(x+t)2+6))+eq\r(\f(17x+28,18x2+6)))),=eq\a\ac(lim,t→0)eq\f(\f([17(x+t)+28](18x2+6)-[18(x+t)2+6](17x+28),(18x2+6)2),2teq\r(\f(17x+28,18x2+6))),=eq\a\ac(lim,t→0)eq\f(t(102-306x2-1008x+306xt+504t),2(18x2+6)2eq\r(\f(17x+28,18x2+6)))=eq