What is the tangent of f(x)= 1/(sqrt(5x)) ?

The tangent of f(x)= 1/(sqrt(5x)) is y=-1/(2sqrt(5)a_{0)^{3/2}}x+3/(2sqrt(5)\sqrt{a_{0)}}

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tangent of f(x)= 1/(sqrt(5x))

tangent of

f (x) = \frac{1}{5 x}

y = - \frac{1}{2 5 a _{0}^{\frac{3}{2}}} x + \frac{3}{2 5 a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{5 a _{0}})

Find the slope of

f (x) = \frac{1}{5 x} : \frac{df ( x )}{d x} = - \frac{1}{2 5 x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{1}{5 x}

x = a_{0} : m = - \frac{1}{2 5 a _{0}^{\frac{3}{2}}}

Find the line with slope m=

- \frac{1}{2 5 a _{0}^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{1}{5 a _{0}}) : y = - \frac{1}{2 5 a _{0}^{\frac{3}{2}}} x + \frac{3}{2 5 a _{0}}

y = - \frac{1}{2 5 a _{0}^{\frac{3}{2}}} x + \frac{3}{2 5 a _{0}}

5 e^{20}

\int (x - 2)^{3} ln (x - 2) d x

(\frac{4}{5} x)^{3}

1 - 2 x^{2}

y^{^{'}} = \frac{( x ^{2} + 1 )}{3 y ^{2}}

What is the tangent of f(x)= 1/(sqrt(5x)) ?
The tangent of f(x)= 1/(sqrt(5x)) is y=-1/(2sqrt(5)a_{0)^{3/2}}x+3/(2sqrt(5)\sqrt{a_{0)}}