How do you implicitly differentiate $xy- yln(x-y)= 4-x$ ?

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Answer 1 · 2018-04-22T04:26:17

$(dy)/dx=(-y-1+y/(x-y))/(x-ln(x-y)+y/(x-y))$

$d/dx(xy-yln(x-y))=d/dx(4-x)$

$y+x(dy)/dx-(dy)/dxln(x-y)-y/(x-y)*d/dx(x-y)=-1$

$y+x(dy)/dx-(dy)/dxln(x-y)-y/(x-y)(1-(dy)/dx)=-1$

$y+x(dy)/dx-(dy)/dxln(x-y)-y/(x-y)+y/(x-y)(dy)/dx=-1$

now isolate $(dy)/dx$

$x(dy)/dx-(dy)/dxln(x-y)+y/(x-y)(dy)/dx=-y-1+y/(x-y)$

$(dy)/dx(x-ln(x-y)+y/(x-y))=-y-1+y/(x-y)$

$(dy)/dx=(-y-1+y/(x-y))/(x-ln(x-y)+y/(x-y))$

How do you implicitly differentiate xy- yln(x-y)= 4-xxy- yln(x-y)= 4-x?