What is the tangent of f(x)=ln(x^2-9x+1),\at x=9 ?

The tangent of f(x)=ln(x^2-9x+1),\at x=9 is y=9x-81

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tangent of f(x)=ln(x^2-9x+1),\at x=9

tangent of

f (x) = ln (x^{2} - 9 x + 1), at x = 9

y = 9 x - 81

Solution steps

Find the tangent point:

(9, 0)

Find the slope of

f (x) = ln (x^{2} - 9 x + 1) : \frac{df ( x )}{d x} = \frac{2 x - 9}{x ^{2} - 9 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 = 9

Find the line with slope m=

9

and passing through

(9, 0) : y = 9 x - 81

y = 9 x - 81

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\int 7 arcsin (x) d x

\frac{d}{d x} ((\frac{x ^{2} + 3}{x ^{2} - 3})^{5})

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- 5 (7 x^{2} + 4)^{- 6}

What is the tangent of f(x)=ln(x^2-9x+1),\at x=9 ?
The tangent of f(x)=ln(x^2-9x+1),\at x=9 is y=9x-81