What is implicit xy^2+4xy-4=0 ?

The answer to implicit xy^2+4xy-4=0 is (dy)/(dx)=(-y^2-4y)/(2x(2+y))

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implicit xy^2+4xy-4=0

implicit derivative

x y^{2} + 4 x y - 4 = 0

\frac{d y}{d x} = \frac{- y ^{2} - 4 y}{2 x ( 2 + y )}

Solution steps

x y^{2} + 4 x y - 4 = 0

Treat

y

y (x)

Differentiate both sides:

2 x y \frac{d y}{d x} + y^{2} + 4 (x \frac{d y}{d x} + y) = 0

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{- y ^{2} - 4 y}{2 x ( 2 + y )}

\frac{d y}{d x} = \frac{- y ^{2} - 4 y}{2 x ( 2 + y )}

\int x^{3} - 4 x^{- 2} + 5 d x

\int_{8}^{24} (x + \frac{1}{x}) d x

\frac{4 t ^{3}}{3} + 6

f (x) = (x^{2} + 1) (x^{3} - x) (3 x^{4} + 2 x - 1)

x \to 0 lim (ln (\frac{- 1}{6} x))

What is implicit xy^2+4xy-4=0 ?
The answer to implicit xy^2+4xy-4=0 is (dy)/(dx)=(-y^2-4y)/(2x(2+y))