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

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

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tangent of y=ln(x^2-2x+1),(2,0)

tangent of

y = ln (x^{2} - 2 x + 1), (2, 0)

y = 2 x - 4

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x - 4

\frac{d}{d x} (2^{s i n (x)})

\int_{0}^{4} \frac{ln ( x )}{x} d x

sin^{2} (x)

\frac{d}{d t} (3 cos (t))

x \to 0 lim (3 ln (x))

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