What is the tangent of y=(((x-1))/((x+1)))^2,\at x=0 ?

The tangent of y=(((x-1))/((x+1)))^2,\at x=0 is y=-4x+1

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tangent of y=(((x-1))/((x+1)))^2,\at x=0

tangent of

y = (\frac{( x - 1 )}{( x + 1 )})^{2}, at x = 0

y = - 4 x + 1

Solution steps

Find the tangent point:

(0, 1)

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}

x = 0 : m = - 4

Find the line with slope m=

- 4

and passing through

(0, 1) : y = - 4 x + 1

y = - 4 x + 1

2 \frac{d y}{d x} + 10 y = 1

\int_{0}^{3} 9 - x^{2} d x

\int \frac{( 3 x + 5 )}{( x ^{2} + 2 x + 2 ) ^{2}} d x

x ln (x) y^{^{'}} + y = x e^{x}, y (2) = - 5

\frac{d}{d x} (5 x x)

What is the tangent of y=(((x-1))/((x+1)))^2,\at x=0 ?
The tangent of y=(((x-1))/((x+1)))^2,\at x=0 is y=-4x+1