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

The tangent of f(x)= 1/(sqrt(x)),\at x=9 is y=-1/54 x+1/2

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

tangent of

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

y = - \frac{1}{54} x + \frac{1}{2}

Solution steps

Find the tangent point:

(9, \frac{1}{3})

Find the slope of

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

Compute the slope of

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

x = 9 : m = - \frac{1}{54}

Find the line with slope m=

- \frac{1}{54}

and passing through

(9, \frac{1}{3}) : y = - \frac{1}{54} x + \frac{1}{2}

y = - \frac{1}{54} x + \frac{1}{2}

\frac{4}{e ^{2 x}}

x \to 0 lim ((\frac{5}{6})^{\frac{1}{x}})

(1 - x^{2})^{- \frac{1}{2}}

\frac{d}{d t} (e^{- t^{2}})

\int x ln^{2} (5 x) d x

What is the tangent of f(x)= 1/(sqrt(x)),\at x=9 ?
The tangent of f(x)= 1/(sqrt(x)),\at x=9 is y=-1/54 x+1/2