What is the tangent of x^2+2xy-y^2+x=6,(2,4) ?

The tangent of x^2+2xy-y^2+x=6,(2,4) is y= 13/4 x-5/2

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tangent of x^2+2xy-y^2+x=6,(2,4)

tangent of

x^{2} + 2 x y - y^{2} + x = 6, (2, 4)

y = \frac{13}{4} x - \frac{5}{2}

Solution steps

Find the slope of

x^{2} + 2 x y - y^{2} + x = 6 : \frac{d y}{d x} = \frac{- 2 x - 1 - 2 y}{2 ( x - y )}

Compute the slope of

x^{2} + 2 x y - y^{2} + x = 6

(2, 4) : m = \frac{13}{4}

Find the line with slope m=

\frac{13}{4}

and passing through

(2, 4) : y = \frac{13}{4} x - \frac{5}{2}

y = \frac{13}{4} x - \frac{5}{2}

\int x^{- 3} (x + 1) d x

\int_{0}^{1} 8^{2 x} d x

\frac{\partial}{\partial y} (x \cdot sin (x + y))

(x^{2} + 6) y^{^{'}} = x y

- 4 y^{^{''}} + 16 y = 0

What is the tangent of x^2+2xy-y^2+x=6,(2,4) ?
The tangent of x^2+2xy-y^2+x=6,(2,4) is y= 13/4 x-5/2