What is the tangent of 1/(1-x) ?

The tangent of 1/(1-x) is y= 1/((1-a_{0))^2}x+(-2a_{0}+1)/((-a_{0)+1)^2}

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tangent of 1/(1-x)

tangent of

\frac{1}{1 - x}

y = \frac{1}{( 1 - a _{0} ) ^{2}} x + \frac{- 2 a _{0} + 1}{( - a _{0} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{1 - a _{0}})

Find the slope of

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

Compute the slope of

y = \frac{1}{1 - x}

x = a_{0} : m = \frac{1}{( 1 - a _{0} ) ^{2}}

Find the line with slope m=

\frac{1}{( 1 - a _{0} ) ^{2}}

and passing through

(a_{0}, \frac{1}{1 - a _{0}}) : y = \frac{1}{( 1 - a _{0} ) ^{2}} x + \frac{- 2 a _{0} + 1}{( - a _{0} + 1 ) ^{2}}

y = \frac{1}{( 1 - a _{0} ) ^{2}} x + \frac{- 2 a _{0} + 1}{( - a _{0} + 1 ) ^{2}}

x \cdot \frac{d y}{d x} = y^{2} - y

x \to 2 lim (4 x^{2} + x - 4)

f (t) = cos (4 t)

arcsin (8 x + 1)

t^{2} (6 t + 5)^{3}

What is the tangent of 1/(1-x) ?
The tangent of 1/(1-x) is y= 1/((1-a_{0))^2}x+(-2a_{0}+1)/((-a_{0)+1)^2}