What is the tangent of f(x)= 1/(sqrt(8x)) ?

The tangent of f(x)= 1/(sqrt(8x)) is y=-1/(4sqrt(2)a_{0)^{3/2}}x+3/(4sqrt(2)\sqrt{a_{0)}}

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tangent of f(x)= 1/(sqrt(8x))

tangent of

f (x) = \frac{1}{8 x}

y = - \frac{1}{4 2 a _{0}^{\frac{3}{2}}} x + \frac{3}{4 2 a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{2 2 a _{0}})

Find the slope of

f (x) = \frac{1}{8 x} : \frac{df ( x )}{d x} = - \frac{1}{4 2 x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{1}{8 x}

x = a_{0} : m = - \frac{1}{4 2 a _{0}^{\frac{3}{2}}}

Find the line with slope m=

- \frac{1}{4 2 a _{0}^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{1}{2 2 a _{0}}) : y = - \frac{1}{4 2 a _{0}^{\frac{3}{2}}} x + \frac{3}{4 2 a _{0}}

y = - \frac{1}{4 2 a _{0}^{\frac{3}{2}}} x + \frac{3}{4 2 a _{0}}

\frac{d}{d x} ((x + 1)^{- 1})

\int (2 x + 2 y) d y

\frac{d}{d x} (ln (\frac{7 x + 8}{4 x + 8}))

y = \frac{( 3 x ^{2} - x - 2 )}{3 x ^{2}}

\int \frac{cos ^{3} ( x )}{sin ^{4} ( x )} d x

What is the tangent of f(x)= 1/(sqrt(8x)) ?
The tangent of f(x)= 1/(sqrt(8x)) is y=-1/(4sqrt(2)a_{0)^{3/2}}x+3/(4sqrt(2)\sqrt{a_{0)}}