What is implicit (dy)/(dx),2x^2+2x+xy=1 ?

The answer to implicit (dy)/(dx),2x^2+2x+xy=1 is (dy}{dx}=\frac{-4x-y-2)/x

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implicit (dy)/(dx),2x^2+2x+xy=1

implicit derivative

\frac{d y}{d x}, 2 x^{2} + 2 x + x y = 1

\frac{d y}{d x} = \frac{- 4 x - y - 2}{x}

Solution steps

2 x^{2} + 2 x + x y = 1

Treat

y

y (x)

Differentiate both sides:

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

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{- 4 x - y - 2}{x}

\frac{d y}{d x} = \frac{- 4 x - y - 2}{x}

f (x) = e^{x} - x^{e}

\frac{\partial}{\partial x} (A^{2} x^{2} + B^{2} y^{2})

\frac{d}{d x} (\frac{lo g _{10} ( x )}{x})

n = 0 \sum \infty \frac{1}{0. 5 ^{n}}

What is implicit (dy)/(dx),2x^2+2x+xy=1 ?
The answer to implicit (dy)/(dx),2x^2+2x+xy=1 is (dy}{dx}=\frac{-4x-y-2)/x