What is the slope of (x-3)^2+(y-1)^2=0 ?

The slope of (x-3)^2+(y-1)^2=0 is (-x+3)/(-1+y)

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slope of (x-3)^2+(y-1)^2=0

slope

(x - 3)^{2} + (y - 1)^{2} = 0

\frac{- x + 3}{- 1 + y}

Solution steps

(x - 3)^{2} + (y - 1)^{2} = 0

Use implicit differentiation to find the slope of

(x - 3)^{2} + (y - 1)^{2} = 0

Implicit Derivative

\frac{d y}{d x}

(x - 3)^{2} + (y - 1)^{2} = 0 : \frac{- x + 3}{- 1 + y}

\frac{- x + 3}{- 1 + y}

d er i v a t i v e y = arcsin (4 x + 1)

a re a y = 2 sin (x), 0 \leq x \leq π

\frac{d y}{d x} = (y - 1)^{2} cos (π x)

d er i v a t i v e g (t) = t^{2} - \frac{4}{t ^{3}}

\frac{\partial}{\partial x} (8 x^{3} y^{2} + 6 x y^{4})

What is the slope of (x-3)^2+(y-1)^2=0 ?
The slope of (x-3)^2+(y-1)^2=0 is (-x+3)/(-1+y)