What is the tangent of f(x)=ln(x^2-2x+1),\at x=2 ?

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

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tangent of f(x)=ln(x^2-2x+1),\at x=2

tangent of

f (x) = ln (x^{2} - 2 x + 1), at x = 2

y = 2 x - 4

Solution steps

Find the tangent point:

(2, 0)

Find the slope of

f (x) = ln (x^{2} - 2 x + 1) : \frac{df ( x )}{d x} = \frac{2}{x - 1}

Compute the slope of

f (x) = ln (x^{2} - 2 x + 1)

x = 2 : m = 2

Find the line with slope m=

2

and passing through

(2, 0) : y = 2 x - 4

y = 2 x - 4

\frac{( x ^{2} )}{2 + x}

\int e^{- 0.06 x} d x

\frac{\partial}{\partial y} (sin (x - y) + (x y))

\frac{\partial}{\partial x} (- 4 x^{2} y - 4 x)

\frac{d}{d x} (\frac{x ^{2} + 4 x + 8}{( x + 2 ) ^{2}})

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