What is the tangent of (x^2+y^2)^3=8x^2y^2,(1,1) ?

The tangent of (x^2+y^2)^3=8x^2y^2,(1,1) is y=-x+2

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tangent of (x^2+y^2)^3=8x^2y^2,(1,1)

tangent of

(x^{2} + y^{2})^{3} = 8 x^{2} y^{2}, (1, 1)

y = - x + 2

Solution steps

Find the slope of

(x^{2} + y^{2})^{3} = 8 x^{2} y^{2} : \frac{d y}{d x} = - \frac{x ( 3 y ^{4} + 6 x ^{2} y ^{2} - 8 y ^{2} + 3 x ^{4} )}{y ( 3 x ^{4} + 6 x ^{2} y ^{2} - 8 x ^{2} + 3 y ^{4} )}

Compute the slope of

(x^{2} + y^{2})^{3} = 8 x^{2} y^{2}

(1, 1) : m = - 1

Find the line with slope m=

- 1

and passing through

(1, 1) : y = - x + 2

y = - x + 2

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

\int_{- 2}^{- 1} \frac{4}{x ^{2}} d x

(13) (1 - 10)

3 e^{x}

\int_{2}^{4} \frac{x ^{2} - 4}{x} d x

What is the tangent of (x^2+y^2)^3=8x^2y^2,(1,1) ?
The tangent of (x^2+y^2)^3=8x^2y^2,(1,1) is y=-x+2