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_{1}^{10} \frac{x ^{2} - 1}{x} d x

\frac{d}{d x} (ln (2 π x^{2}))

\int \frac{e ^{6 x}}{e ^{6 x} sin ^{2} ( 4 x )} d x

x \to 2 lim ((\frac{1}{x})^{3})

\frac{d}{d x} (\frac{3}{8} x^{\frac{2}{3}})

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