What is the tangent of x^2+2xy-y^2+x=12,(3,6) ?

The tangent of x^2+2xy-y^2+x=12,(3,6) is y= 19/6 x-7/2

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tangent of x^2+2xy-y^2+x=12,(3,6)

tangent of

x^{2} + 2 x y - y^{2} + x = 12, (3, 6)

y = \frac{19}{6} x - \frac{7}{2}

Solution steps

Find the slope of

x^{2} + 2 x y - y^{2} + x = 12 : \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 = 12

(3, 6) : m = \frac{19}{6}

Find the line with slope m=

\frac{19}{6}

and passing through

(3, 6) : y = \frac{19}{6} x - \frac{7}{2}

y = \frac{19}{6} x - \frac{7}{2}

\frac{3}{s} + 8 \frac{1}{s ( s + 1 )}

\int sin (x) - 2 cos (x) d x

\int \frac{72}{x ^{2} 4 x ^{2} + 9} d x

n = 0 \sum \infty (5 x)^{n}

\frac{1}{1 - x ^{2}}

What is the tangent of x^2+2xy-y^2+x=12,(3,6) ?
The tangent of x^2+2xy-y^2+x=12,(3,6) is y= 19/6 x-7/2