What is the tangent of f(x)= 1/(sqrt(9x)),\at x=9 ?

The tangent of f(x)= 1/(sqrt(9x)),\at x=9 is y=-1/162 x+1/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of f(x)= 1/(sqrt(9x)),\at x=9

tangent of

f (x) = \frac{1}{9 x}, at x = 9

y = - \frac{1}{162} x + \frac{1}{6}

Solution steps

Find the tangent point:

(9, \frac{1}{9})

Find the slope of

f (x) = \frac{1}{9 x} : \frac{df ( x )}{d x} = - \frac{1}{6 x ^{\frac{3}{2}}}

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 = - \frac{1}{162}

Find the line with slope m=

- \frac{1}{162}

and passing through

(9, \frac{1}{9}) : y = - \frac{1}{162} x + \frac{1}{6}

y = - \frac{1}{162} x + \frac{1}{6}

What is the tangent of f(x)= 1/(sqrt(9x)),\at x=9 ?
The tangent of f(x)= 1/(sqrt(9x)),\at x=9 is y=-1/162 x+1/6