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

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

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