y=ax+b/cx+d,求y',要过程及讲解

显示全部楼层 · 2011-11-3 10:13:07

除法求导法则y=f(x)/g(x)y' = (f ' (x)g(x) - f(x)g ' (x))/g(x)^2（为看清楚哪个带导数中间空开了）所以y=ax+b/cx+dy'=(ax+b)'(cx+d)-(ax+b)(cx+d)' / (cx+d)^2=a(cx+d)-c(ax+b) / (cx+d)^2=(ad-bc) / (cx+d)^2...

千问 · 2011-11-3 10:13:07

y=ax+b/cx+d,y'=(ax+b)'*(cx+d)-(ax+b)*(cx+d)'/(cx+d)^2=(a*(cx+d)-(ax+b)*c)/(cx+d)^2=(ad-bc)/(cx+d)^2...

千问 · 2011-11-3 10:13:07

y'=[(ax＋b)'(cx＋d)－(ax＋b)(cx＋d)']/(cx＋d)2 =[ad－bc]/(cx＋d)2...

千问 · 2011-11-3 10:13:07

导数的公式除法公式！！！参考下y'=[(ax＋b)'(cx＋d)－(ax＋b)(cx＋d)']/(cx＋d)2 =[ad－bc]/(cx＋d)2...

千问 · 2011-11-3 10:13:07

y'=a+b/c
除以x就行了...