How do you differentiate $5 y - x = \frac{x y}{ln (x - y)}$ $5y-x=(xy)/ln(x-y)$ ?

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Answer 1 · 2016-04-14T21:11:28

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

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

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

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

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

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

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

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