What is the tangent of y=ln(x^2-9x+1),(9,0) ?

The tangent of y=ln(x^2-9x+1),(9,0) is y=9x-81

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=ln(x^2-9x+1),(9,0)

tangent of

y = ln (x^{2} - 9 x + 1), (9, 0)

y = 9 x - 81

Solution steps

Find the slope of

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

Compute the slope of

y = ln (x^{2} - 9 x + 1)

(9, 0) : m = 9

Find the line with slope m=

9

and passing through

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

y = 9 x - 81

\int 5 sin (t) d t

\frac{0.625}{s + 4}

x \to 0 lim (\frac{x ^{2} + 2 x - 99}{x})

\int \frac{2 x}{1 + x ^{4}} d x

y = (x + 3)^{3}

What is the tangent of y=ln(x^2-9x+1),(9,0) ?
The tangent of y=ln(x^2-9x+1),(9,0) is y=9x-81