How do you implicitly differentiate #3y + y^4/x^2 = 2#?

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Answer 1 · 2016-07-07T14:33:34

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

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

[ie differentiate wrt x]

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

#implies 3y' + (y^4/x^2)' = 0 qquad triangle#

two of those are very easy. we'll do the middle one using the quotient rule separately

# (y^4/x^2)'#

#= ((y^4)^prime x^2 - y^4(x^2)^prime)/(x^4)#

#= (4y^3 y' x^2 - y^4(2x))/(x^4)# which we stuff straight back into #triangle# to get

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

some algebra

#3y'x^4 +4y^3 y' x^2 - 2x y^4= 0#

#3y'x^3 +4y^3 y' x - 2 y^4= 0#

collecting #y'# s together

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

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