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

The tangent of 1/((x-1)),\at x=2 is y=-x+3

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tangent of 1/((x-1)),\at x=2

tangent of

\frac{1}{( x - 1 )}, at x = 2

y = - x + 3

Solution steps

Find the tangent point:

(2, 1)

Find the slope of

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

Compute the slope of

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

x = 2 : m = - 1

Find the line with slope m=

- 1

and passing through

(2, 1) : y = - x + 3

y = - x + 3

x \to 1 + lim (\frac{x}{x ^{2} - 1})

\frac{\partial}{\partial v} ((u^{2} v - v^{3})^{5})

(- 3, 0), (0, - 3)

\frac{d}{d x} (x \cdot cos^{3} (4 x))

\frac{d}{d x} (x^{4} + e^{- x})

What is the tangent of 1/((x-1)),\at x=2 ?
The tangent of 1/((x-1)),\at x=2 is y=-x+3