What is the tangent of f(x)= 1/(sqrt(x)),\at x=81 ?

The tangent of f(x)= 1/(sqrt(x)),\at x=81 is y=-1/1458 x+1/6

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tangent of f(x)= 1/(sqrt(x)),\at x=81

tangent of

f (x) = \frac{1}{x}, at x = 81

y = - \frac{1}{1458} x + \frac{1}{6}

Solution steps

Find the tangent point:

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

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 = 81 : m = - \frac{1}{1458}

Find the line with slope m=

- \frac{1}{1458}

and passing through

(81, \frac{1}{9}) : y = - \frac{1}{1458} x + \frac{1}{6}

y = - \frac{1}{1458} x + \frac{1}{6}

arccos (x) + x, at x = 0

y = 9 + 5 x^{2} - 2 x^{3}

2 y^{^{'}} + 4 y = 2 t e^{- 2 t}

\int \frac{3 - x}{3 + x} d x

\int \frac{ln ∣ x ∣}{x} d x

What is the tangent of f(x)= 1/(sqrt(x)),\at x=81 ?
The tangent of f(x)= 1/(sqrt(x)),\at x=81 is y=-1/1458 x+1/6