What is derivative of x^2+y^2-2x=1 ?

The answer to derivative of x^2+y^2-2x=1 is (dy}{dx}=\frac{-x+1)/y

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derivative of x^2+y^2-2x=1

derivative of

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

\frac{d y}{d x} = \frac{- x + 1}{y}

Solution steps

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

Treat

y

y (x)

Differentiate both sides:

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

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{- x + 1}{y}

\frac{d y}{d x} = \frac{- x + 1}{y}

p = 7 q^{3} (5 q - 150)

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

e^{x} y = 2 y

y^{3} - 3 x y^{2} = x^{3} + 3 x^{2} y

\frac{x ^{2}}{18} + \frac{y ^{2}}{8} = 1

What is derivative of x^2+y^2-2x=1 ?
The answer to derivative of x^2+y^2-2x=1 is (dy}{dx}=\frac{-x+1)/y