What is the tangent of y=ln(x^2-8x+1),(8,0) ?

The tangent of y=ln(x^2-8x+1),(8,0) is y=8x-64

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tangent of y=ln(x^2-8x+1),(8,0)

tangent of

y = ln (x^{2} - 8 x + 1), (8, 0)

y = 8 x - 64

Solution steps

Find the slope of

y = ln (x^{2} - 8 x + 1) : \frac{d y}{d x} = \frac{2 x - 8}{x ^{2} - 8 x + 1}

Compute the slope of

y = ln (x^{2} - 8 x + 1)

(8, 0) : m = 8

Find the line with slope m=

8

and passing through

(8, 0) : y = 8 x - 64

y = 8 x - 64

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What is the tangent of y=ln(x^2-8x+1),(8,0) ?
The tangent of y=ln(x^2-8x+1),(8,0) is y=8x-64