Calculus and AnalysisI have it down to
3 years ago
I have it down to ln(y)+(x/y)y'=y'ln(x)+y/x The problem is factoring out y', which leads to either y'=(ln(y)-(y/x))/(ln(x)-(x/y)) or to y'=((y/x)-ln(y))/((x/y)-ln(x)) Am I missing something?
2 Answers
Best Answer
briannaciamaichelo Staff answered 3 years ago
x^y=y^x -> ylog x=xlog y -> log xdy+(y)/(x)dx=log ydx+(x)/(y)dy -> (log x-(x)/(y))dy=(log y-(y)/(x))dx -> (dy)/(dx)=(log y-(y)/(x))/(log x-(x)/(y))
Best Answer
Ben Staff answered 3 years ago
Write e^(yln x)=e^(xln y) Then ((y)/(x)+y'ln x)e^(yln x)=(ln y+(xy')/(y))e^(xln y) rearranging, we get (x^yln x-(x)/(y)y^x)y'=y^xln y-(y)/(x)x^y Use y^x=x^y and get y'=(ln y-(y/x))/(ln x-(x/y))
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