How do you differentiate #5y-x=(xy)/ln(x-y)#?

1 Answer
Bdub
Apr 14, 2016

#y'=[yln(x-y)-(xy)/(x-y)+(ln(x-y))^2]/[5(ln(x-y))^2-xln(x-y)-(xy)/(x-y)]#

Explanation:

#5y'-1=(ln(x-y)[xy'+y]-xy(1/(x-y)(1-y')))/(ln(x-y))^2#

#(5y'-1)xx(ln(x-y))^2=ln(x-y)[xy'+y]-(xy)/(x-y)(1-y'))#

#5y'(ln(x-y))^2-(ln(x-y))^2=xy'ln(x-y)+yln(x-y)-(xy)/(x-y)+(xyy')/(x-y)#

#5y'(ln(x-y))^2-xy'ln(x-y)-(xyy')/(x-y) = yln(x-y)-(xy)/(x-y)+(ln(x-y))^2#

#y'[5(ln(x-y))^2-xln(x-y)-(xy)/(x-y)]=yln(x-y)-(xy)/(x-y)+(ln(x-y))^2#

#y'=[yln(x-y)-(xy)/(x-y)+(ln(x-y))^2]/[5(ln(x-y))^2-xln(x-y)-(xy)/(x-y)]#

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