Question

隐函数e^y*y'-sin(x^2+y)*(2x+y')=0如何化成y'=的形式啊..能写出过程么,

Answer 1

解出来就行了啊

e^y*y'-sin(x^2+y)*(2x+y')=0

e^y*y'-sin(x^2+y)*2x-sin(x^2+y)*y'=0

e^y*y'-sin(x^2+y)*y'=sin(x^2+y)*2x

y'=sin(x^2+y)*2x/[e^y-sin(x^2+y)]