What is the tangent of x^2+2xy-y^2+x=5,(3,7) ?

The tangent of x^2+2xy-y^2+x=5,(3,7) is y= 21/8 x-7/8

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tangent of x^2+2xy-y^2+x=5,(3,7)

tangent of

x^{2} + 2 x y - y^{2} + x = 5, (3, 7)

y = \frac{21}{8} x - \frac{7}{8}

Solution steps

Find the slope of

x^{2} + 2 x y - y^{2} + x = 5 : \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 = 5

(3, 7) : m = \frac{21}{8}

Find the line with slope m=

\frac{21}{8}

and passing through

(3, 7) : y = \frac{21}{8} x - \frac{7}{8}

y = \frac{21}{8} x - \frac{7}{8}

\frac{d}{d t} (t ln (t))

x \to 0 + lim (\frac{x}{ln ( x )})

f (x) = x^{2} + 4 x - 5, - 3, 3

x \to 3 lim (\frac{5 x}{x - 3})

\frac{d}{d x} (ln (\frac{3 x}{2 x + 1}))

What is the tangent of x^2+2xy-y^2+x=5,(3,7) ?
The tangent of x^2+2xy-y^2+x=5,(3,7) is y= 21/8 x-7/8