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

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Answer 1 · 2018-07-12T19:34:38

#y'=(3x^2*(x+4y)^2+y)/x#

Proof:

#(y'(x+4y)-y(1+4y'))/(x+4y)^2=3x^2#
multiplying by #(x+4y)^2#

#y'(x+4y)-y-4yy'=3x^2(x+4y)^2#

adding #y#

#y'(x+4y-4y)=3x^2(x+4y)^2+y#

so

#y'=(3x^2(x+4y)^2+y)/x# for #xne 0#