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

The tangent of f(x)= 2/(sqrt(x)),\at (1/9) is y=-1/(a_{0)^{3/2}}x+3/(sqrt(a_{0))}

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

tangent of

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

y = - \frac{1}{a _{0}^{\frac{3}{2}}} x + \frac{3}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{2}{a _{0}})

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 = a_{0} : m = - \frac{1}{a _{0}^{\frac{3}{2}}}

Find the line with slope m=

- \frac{1}{a _{0}^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{2}{a _{0}}) : y = - \frac{1}{a _{0}^{\frac{3}{2}}} x + \frac{3}{a _{0}}

y = - \frac{1}{a _{0}^{\frac{3}{2}}} x + \frac{3}{a _{0}}

\int x arctan (4 x) d x

x^{2} - 4 x + 1

\int sin (x) cos^{6} (x) d x

x \to 0 lim (cos^{2} (x))

\frac{x ^{2} + 1}{x - 2}

What is the tangent of f(x)= 2/(sqrt(x)),\at (1/9) ?
The tangent of f(x)= 2/(sqrt(x)),\at (1/9) is y=-1/(a_{0)^{3/2}}x+3/(sqrt(a_{0))}