三种方法计算形如函数y=√(ax+b).(cx^2+d)一阶导数过程步骤B17

函数y=√[(8x+30)/(15x2+28)]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+30)/(15x2+28)]的一阶导数的具体步骤。思路一：幂函数导数法∵y=√[(8x+30)/(15x2+28)]=[(8x+30)/(15x2+28)]^(1/2),∴y'=(1/2)*[(8x+30)/(15x2+28)]^(-1/2)*[8(15x2+28)-30x(8x+30)]/(15x2+28)2=(-1/2)*[(120x2+900x-224)/√[(8x+30)(15x2+28)]3。思路二：取对数求导数法∵y=√[(8x+30)/(15x2+28)]，∴lny=(1/2)ln[(8x+30)/ln(15x2+28)],即：2lny=ln(8x+30)-ln(15x2+28),两边同时对x求导得：2y'/y=8/(8x+30)-30x/(15x2+28)。进一步化简有y'=1/2·y·[8(15x2+28)-30x(8x+30)]/[(8x+30)(15x2+28)]=-1/2·y·(120x2+900x-224)/[(8x+30)(15x2+28)],=-1/2·√[(8x+30)/(15x2+28)]·(120x2+900x-224)/[(8x+30)(15x2+28)],=1/2·(120x2+900x-224)/[3√((8x+30)(15x2+28)]3.思路三：定义法求解导数∵f(x)=√[(8x+30)/(15x2+28)],∴y'=lim(t→0)[f(x+t)-f(x)]/t=lim(t→0)[√[(8x+8t+30)/(15x(x+t)2+28)-√[(8x+30)/(15x2+28)]/t=lim(t→0)[t(224-120x2-900x+120xt+450t)]/[2(15x2+28)2√[(8x+30)/(15x2+28)].y'=(224-120x2-900x+0)/{2(15x2+28)2√[(8x+30)/(15x2+28)]}=(-1/2)·(120x2+900x-224)/{√[(8x+30)(15x2+28)]3}。函数y=eq\r(\f(8x+30,15x2+28))一阶导数三种求法思路一：幂函数导数法y'=eq\f(1,2)·eq(\f(15x2+28,8x+30))\s\up12(\f(1,2))·eq\f(8(15x2+28)-30x(8x+30),(15x2+28)2),=eq\f(1,2)·eq(\f(15x2+28,8x+30))\s\up12(\f(1,2))·eq\f(224-120x2-900x,(15x2+28)2)=eq-\f(1,2)·eq\f(120x2+900x-224,\r((8x+30)(15x2+28)3))。思路二：取对数求导数法∵y=eq\r(\f(8x+30,15x2+28))，∴lny=eq\f(1,2)eqln\f(8x+30,15x2+28),即：2lny=ln(8x+30)-ln(15x2+28),两边同时对x求导得：eq\f(2y',y)=eq\f(8,8x+30)-eq\f(30x,15x2+28),则：y'=eq\f(1,2)·y·eq\f(8(15x2+28)-30x(8x+30),(8x+30)(15x2+28)),=-eq\f(1,2)·y·eq\f(120x2+900x-224,(8x+30)(15x2+28)),=-eq\f(1,2)·eq\r(\f(8x+30,15x2+28))·eq\f(120x2+900x-224,(8x+30)(15x2+28)),=-eq\f(1,2)·eq\f(120x2+900x-224,\r((8x+30)(15x2+28)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)+30,15(x+t)2+28))-eq\r(\f(8x+30,15x2+28)),t),=eq\a\ac(lim,t→0)eq\f(eq\f(8(x+t)+30,15(x+t)2+28)-eq\f(8x+30,15x2+28),t(eq\r(\f(8(x+t)+30,15(x+t)2+28))+eq\r(\f(8x+30,15x2+28)))),=eq\a\ac(lim,t→0)eq\f(\f([8(x+t)+30](15x2+28)-[15(x+t)2+28](8x+30),[15(x+t)2+28](15x2+28)),t(eq\r(\f(8(x+t)+30,15(x+t)2+28))+eq\r(\f(8x+30,15x2+28)))),=eq\a\ac(lim,t→0)eq\f(\f([8(x+t)+30](15x2+28)-[15(x+t)2+28](8x+30),(15x2+28)2),2teq\r(\f(8x+30,15x2+28))),=eq\a\ac(lim,t→0)eq\f(t(224-120x2-900x+120xt+450t),2(15x2+28)2eq\r(\f(8x+30,15x2+28)))=eq