What is the tangent of 1/(sqrt(9x)) ?

The tangent of 1/(sqrt(9x)) is y=-1/(6a_{0)^{3/2}}x+1/(2sqrt(a_{0))}

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tangent of 1/(sqrt(9x))

tangent of

\frac{1}{9 x}

y = - \frac{1}{6 a _{0}^{\frac{3}{2}}} x + \frac{1}{2 a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{3 a _{0}})

Find the slope of

y = \frac{1}{9 x} : \frac{d y}{d x} = - \frac{1}{6 x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{1}{9 x}

x = a_{0} : m = - \frac{1}{6 a _{0}^{\frac{3}{2}}}

Find the line with slope m=

- \frac{1}{6 a _{0}^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{1}{3 a _{0}}) : y = - \frac{1}{6 a _{0}^{\frac{3}{2}}} x + \frac{1}{2 a _{0}}

y = - \frac{1}{6 a _{0}^{\frac{3}{2}}} x + \frac{1}{2 a _{0}}

\int \frac{37 x + 5}{( 6 x + 1 ) ( x - 1 )} d x

f (x) = x \cdot ln (x)

\frac{\partial}{\partial x} (x^{8} + 6 x y^{7})

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

y^{^{''}} + 4 y = 3 cos (2 t) - 5 sin (2 t)

What is the tangent of 1/(sqrt(9x)) ?
The tangent of 1/(sqrt(9x)) is y=-1/(6a_{0)^{3/2}}x+1/(2sqrt(a_{0))}