What is the tangent of (x^2+y^2)^2=4x^2y,(1,1) ?

The tangent of (x^2+y^2)^2=4x^2y,(1,1) is y=1

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tangent of (x^2+y^2)^2=4x^2y,(1,1)

tangent of

(x^{2} + y^{2})^{2} = 4 x^{2} y, (1, 1)

y = 1

Solution steps

Find the slope of

(x^{2} + y^{2})^{2} = 4 x^{2} y : \frac{d y}{d x} = \frac{x ( - y ^{2} + 2 y - x ^{2} )}{- x ^{2} + y ^{3} + x ^{2} y}

Compute the slope of

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

(1, 1) : m = 0

Find the line with slope m=

0

and passing through

(1, 1) : y = 1

y = 1

\int \frac{- 12}{x ^{2} 4 - x ^{2}} d x

\frac{\partial}{\partial x} (4 x^{2} + y^{2} - 7 y)

\frac{d}{d x} (\frac{6 x}{x ^{2} + 3})

\frac{d}{d x} (\frac{1}{( x ^{3} + 9 ) ^{5}})

\frac{d}{d x} (sec^{2} (3 - 2 x))

What is the tangent of (x^2+y^2)^2=4x^2y,(1,1) ?
The tangent of (x^2+y^2)^2=4x^2y,(1,1) is y=1