What is the tangent of y=((x+1)/(x-1))^2,(3,4) ?

The tangent of y=((x+1)/(x-1))^2,(3,4) is y=-2x+10

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tangent of y=((x+1)/(x-1))^2,(3,4)

tangent of

y = (\frac{x + 1}{x - 1})^{2}, (3, 4)

y = - 2 x + 10

Solution steps

Find the slope of

y = (\frac{x + 1}{x - 1})^{2} : \frac{d y}{d x} = - \frac{4 ( x + 1 )}{( x - 1 ) ^{3}}

Compute the slope of

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

(3, 4) : m = - 2

Find the line with slope m=

- 2

and passing through

(3, 4) : y = - 2 x + 10

y = - 2 x + 10

f (x) = x^{9}, at x = - 1

\frac{d}{d x} (e^{2 x} n (3 x))

\frac{3}{8} π r^{3}

x \to 0 lim (8 csc (x) - \frac{8}{x})

n = 0 \sum \infty 1^{- n}

What is the tangent of y=((x+1)/(x-1))^2,(3,4) ?
The tangent of y=((x+1)/(x-1))^2,(3,4) is y=-2x+10