How do you implicitly differentiate $- y = x^{3} y^{2} - 3 x^{2} y^{3} - 7 x y^{4}$ $-y= x^3y^2-3x^2y^3-7xy^4$ ?

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Answer 1 · 2016-02-17T01:33:32

A useful timesaver is to use $y'$ instead of $dy/dx$ when differentiating y ... here is how it works:

$-y= x^3y^2-3x^2y^3-7xy^4$

Now, using the chain and product rules ...

$-y'= 3x^2y^2+x^3(2yy')-6xy^3-3x^2(3y^2y')-7y^4-7x(4y^3y')$

Next combine the $y'$ terms on the left side of the equation ...

$y'[-1-x^3(2y)+3x^2(3y^2)+7x(4y^3)]=3x^2y^2-6xy^3-7y^4$

Finally, simplify and solve for $y'$

$y'=(3x^2y^2-6xy^3-7y^4)/[9x^2(y^2)+28x(y^3)-x^3(2y)-1]$

hope that helped

How do you implicitly differentiate -y= x^3y^2-3x^2y^3-7xy^4 −y=x3y2−3x2y3−7xy4-y= x^3y^2-3x^2y^3-7xy^4 ?