How do you implicitly differentiate #(x/(x-4y))=x^(3)/y^2-1#?

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Answer 1 · 2016-07-18T15:25:03

#y'=(2x^3-y^2-6x^2y)/(2xy+2x^3-6y^2)#

#(x/(x-4y))=x^3/y^2-1=(x^3-y^2)/y^2#

#:.xy^2=(x-4y)(x^3-y^2)=x^4-xy^2-4x^3y+4y^3#

#:.xy^2+xy^2+4x^3y=x^4+4y^3#

#:.2xy^2+4x^3y=x^4+4y^3#

Diff.ing both sides w.r.t. #x#,

#d/dx(2xy^2)+d/dx(4x^3y)=d/dx(x^4)+d/dx(4y^3)#

#:. 2{xd/dx(y^2)+y^2d/dx(x)}+4{x^3dy/dx+3x^2y}#

#=4x^3+4d/dx(y^3)........(1)#

Here, let us note that, by Chain Rule,

#d/dx(y^2)=d/dy(y^2)*dy/dx=2y*dy/dx=2yy'#,

and on the same lines, #d/dx(y^3)=3y^2y'#.

These results will be used in #(1)# [after cancelling #2# throughout]

to get,

#{x*2yy'+y^2*1}+2{x^3y'+3x^2y}=2x^3+2*3y^2y'#

#2xyy'+y^2+2x^3y'+6x^2y=2x^3+6y^2y'#

#y'(2xy+2x^3-6y^2)=2x^3-y^2-6x^2y#

#y'=(2x^3-y^2-6x^2y)/(2xy+2x^3-6y^2)#