What is the tangent of f(x)= 2/(sqrt(x)),\at x= 1/9 ?

The tangent of f(x)= 2/(sqrt(x)),\at x= 1/9 is y=-27x+9

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tangent of f(x)= 2/(sqrt(x)),\at x= 1/9

tangent of

f (x) = \frac{2}{x}, at x = \frac{1}{9}

y = - 27 x + 9

Solution steps

Find the tangent point:

(\frac{1}{9}, 6)

Find the slope of

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

Compute the slope of

f (x) = \frac{2}{x}

x = \frac{1}{9} : m = - 27

Find the line with slope m=

- 27

and passing through

(\frac{1}{9}, 6) : y = - 27 x + 9

y = - 27 x + 9

t \to + 0 lim (\frac{sin ( 7 t )}{t})

\frac{d}{d x} (ln (\frac{1}{2}))

f (x) = - 2 x^{\frac{2}{3}} - 4, at x = 8

\int \frac{x ^{3}}{1 - x ^{8}} d x

What is the tangent of f(x)= 2/(sqrt(x)),\at x= 1/9 ?
The tangent of f(x)= 2/(sqrt(x)),\at x= 1/9 is y=-27x+9