How do you implicitly differentiate #22=(y)/(1-xe^y)#?

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Answer 1 · 2018-04-07T19:00:23

#(dy)/(dx)=(22e^y)/(y-23)#

We have;

#22=y/(1-xe^y)#

#=>22-22xe^y=y#

#22-y=22xe^y#

#(22-y)/(22e^y)=x#

#x=(22-y)/(22e^y)#

Diff.w.r.t. #color(red)(y)# and #"using"color(blue)" Quotient Rule"#

#(dx)/(dy)=(22e^yd/(dy)(22-y)-(22-y)d/(dy)(22e^y))/(22e^y)^2#

#=>(dx)/(dy)=(22e^y(-1)-(22-y)(22e^y))/(22e^y)^2#

#=>(dx)/(dy)=(22e^y(-1-22+y))/(22e^y)^2#

#=>color(red)((dx)/(dy))=(y-23)/(22e^y)to#,But , #[(dx)/(dy)=1/((dy)/(dx))]#

#=>color(red)((dy)/(dx))=(22e^y)/(y-23)#