What is the implicit derivative of #1= x/y^3-ye^(x-y)#?

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Answer 1 · 2018-04-14T10:10:02

#(dy)/(dx)=(1-y^4e^(x-y))/[3y^2-y^4e^(x-y)+4e^(x-y)y^3]#

Here,

#1=x/y^3-ye^(x-y)#

#=>y^3=x-y^4e^(x-y)#

Diff.w.r.t. #x#, #"using "color(blue) "product and chain rule"#

#3y^2(dy)/(dx)=1-[y^4d/(dx)(e^(x-y))+e^(x-y)d/(dx)(y^4)]#

#3y^2(dy)/(dx)=1-[y^4e^(x-y)(1-(dy)/(dx))+e^(x-y)4y^3(dy)/(dx)]#

#3y^2(dy)/(dx)=1-y^4e^(x-y)+y^4e^(x-y)(dy)/(dx)-4e^(x- y)y^3(dy)/(dx)#

#3y^2(dy)/(dx)-y^4e^(x-y)(dy)/(dx)+4e^(x-y)y^3(dy)/(dx)=1- y^4e^(x-y)#

#(dy)/(dx)[3y^2-y^4e^(x-y)+4e^(x-y)y^3]=1-y^4e^(x-y)#

#(dy)/(dx)=(1-y^4e^(x-y))/[3y^2-y^4e^(x-y)+4e^(x-y)y^3]#