對(duì)于一個(gè)隱函數(shù)
要求導(dǎo)數(shù)dy/dx
只要在這函數(shù)兩邊同時(shí)對(duì)x求導(dǎo)就可以了
最后再整理出dy/dx
如果有個(gè)實(shí)例就更好說(shuō)了
有不懂歡迎追問(wèn)我沒有輸完就提問(wèn)了方程xy=e^(x+y)我是這樣做的lnxy = x+y;lnx + lny = x+y兩邊取導(dǎo)數(shù)1/x+1/y*(dy/dx) = 1+dy/dxdy/dx = ((1-x)y)/((1-y)x)這樣做對(duì)不xy=e^(x+y)同對(duì)x求導(dǎo)：y+xy'=e^(x+y)*(1+y')xy'-e^(x+y)y'=e^(x+y)-y(x-e^(x+y))y'=(e^(x+y)-y)y'=(e^(x+y)-y) / (x-e^(x+y))因此，dy/dx=(e^(x+y)-y) / (x-e^(x+y))=(xy-y) / (x-xy)=y(x-1) / x(1-y) 所以其實(shí)兩種做法都對(duì)相比之下，你的做法，比我的更簡(jiǎn)單~~有不懂歡迎追問(wèn)其實(shí)我知道有這2種做法，我想問(wèn)為什么結(jié)果會(huì)不一樣xy=e^(x+y)同取對(duì)數(shù)：ln(xy)=(x+y)lnelnx+lny=x+y同對(duì)x求導(dǎo)：1/x+y'/y=1+y'1/x-1=(1-1/y)y'y'=(1/x-1)/(1-1/y) =(1-x)/x / (y-1)/y =y(1-x) / x(y-1) =y(x-1) / x(1-y)兩種方法都是一樣的，只是你算的時(shí)候有個(gè)負(fù)號(hào)漏掉而已~~~有不懂歡迎追問(wèn)