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

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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}}}

Compute the slope of

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

x = 9 : 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}

\int cos (\frac{π}{x}) d x

f (x) = x (2 x + 2)

x \to 1 lim (\frac{5 - x ^{2}}{1 + x})

\int 0.1

f (x) = 4 e^{x^{- 3}}

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