What is the tangent of 256=4x^2+2xy-4y^3,(2,-4) ?

The tangent of 256=4x^2+2xy-4y^3,(2,-4) is y= 2/47 x-192/47

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tangent of 256=4x^2+2xy-4y^3,(2,-4)

tangent of

256 = 4 x^{2} + 2 x y - 4 y^{3}, (2, - 4)

y = \frac{2}{47} x - \frac{192}{47}

Solution steps

Find the slope of

256 = 4 x^{2} + 2 x y - 4 y^{3} : \frac{d y}{d x} = - \frac{4 x + y}{x - 6 y ^{2}}

Compute the slope of

256 = 4 x^{2} + 2 x y - 4 y^{3}

(2, - 4) : m = \frac{2}{47}

Find the line with slope m=

\frac{2}{47}

and passing through

(2, - 4) : y = \frac{2}{47} x - \frac{192}{47}

y = \frac{2}{47} x - \frac{192}{47}

\int \frac{cos ( 3 x )}{sin ( 3 x )} d x

\int \frac{cos ( 2 x )}{cos ^{2} ( x ) sin ^{2} ( x )} d x

\int \frac{1}{( u ^{2} + a ^{2} ) ^{2}} d u

n = 3 \sum \infty (\frac{- 1}{2})^{n}

y^{^{''}} + 10 y^{^{'}} + 26 y = 0

What is the tangent of 256=4x^2+2xy-4y^3,(2,-4) ?
The tangent of 256=4x^2+2xy-4y^3,(2,-4) is y= 2/47 x-192/47