What is the tangent of 1/(sqrt(5x)),\at x=9 ?

The tangent of 1/(sqrt(5x)),\at x=9 is y=-1/(54sqrt(5))x+1/(2sqrt(5))

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tangent of 1/(sqrt(5x)),\at x=9

tangent of

\frac{1}{5 x}, at x = 9

y = - \frac{1}{54 5} x + \frac{1}{2 5}

Solution steps

Find the tangent point:

(9, \frac{1}{3 5})

Find the slope of

y = \frac{1}{5 x} : \frac{d y}{d x} = - \frac{1}{2 5 x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{1}{5 x}

x = 9 : m = - \frac{1}{54 5}

Find the line with slope m=

- \frac{1}{54 5}

and passing through

(9, \frac{1}{3 5}) : y = - \frac{1}{54 5} x + \frac{1}{2 5}

y = - \frac{1}{54 5} x + \frac{1}{2 5}

\frac{1}{( ( 3 - 2 x ) )}

\frac{5 x + 2}{6 x - 4}

y = x, (0, 6)

\int_{0}^{4} 2 t + 4 d t

n = 0 \sum \infty \frac{1}{( n ( n + 2 ) )}

What is the tangent of 1/(sqrt(5x)),\at x=9 ?
The tangent of 1/(sqrt(5x)),\at x=9 is y=-1/(54sqrt(5))x+1/(2sqrt(5))