What is the tangent of f(x)= 6/(sqrt(x)) ?

The tangent of f(x)= 6/(sqrt(x)) is y=-3/(a_{0)^{3/2}}x+9/(sqrt(a_{0))}

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tangent of f(x)= 6/(sqrt(x))

tangent of

f (x) = \frac{6}{x}

y = - \frac{3}{a _{0}^{\frac{3}{2}}} x + \frac{9}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{6}{a _{0}})

Find the slope of

f (x) = \frac{6}{x} : \frac{df ( x )}{d x} = - \frac{3}{x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{6}{x}

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

Find the line with slope m=

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

and passing through

(a_{0}, \frac{6}{a _{0}}) : y = - \frac{3}{a _{0}^{\frac{3}{2}}} x + \frac{9}{a _{0}}

y = - \frac{3}{a _{0}^{\frac{3}{2}}} x + \frac{9}{a _{0}}

\int 3 t d t

e^{- 2 t} \cdot (t^{2} + 4 t + 5)

\frac{\partial}{\partial x} \frac{1}{f ( x ) ^{^{'}} ( f ( x ) ^{- 1} ( x ) )}

\frac{\partial}{\partial x} (\frac{y}{2 x} - 1)

\frac{\partial}{\partial x} (2 π \frac{x}{y})

What is the tangent of f(x)= 6/(sqrt(x)) ?
The tangent of f(x)= 6/(sqrt(x)) is y=-3/(a_{0)^{3/2}}x+9/(sqrt(a_{0))}