What is the tangent of 1/(sqrt(10x)),\at x=9 ?

The tangent of 1/(sqrt(10x)),\at x=9 is y=-1/(54sqrt(10))x+1/(2sqrt(10))

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tangent of 1/(sqrt(10x)),\at x=9

tangent of

\frac{1}{10 x}, at x = 9

y = - \frac{1}{54 10} x + \frac{1}{2 10}

Solution steps

Find the tangent point:

(9, \frac{1}{3 10})

Find the slope of

y = \frac{1}{10 x} : \frac{d y}{d x} = - \frac{1}{2 10 x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{1}{10 x}

x = 9 : m = - \frac{1}{54 10}

Find the line with slope m=

- \frac{1}{54 10}

and passing through

(9, \frac{1}{3 10}) : y = - \frac{1}{54 10} x + \frac{1}{2 10}

y = - \frac{1}{54 10} x + \frac{1}{2 10}

\frac{d}{d x} (\frac{x ^{3}}{x ^{2} - 9})

\int ln (x) d x

\int \frac{t ^{3} - 8 t + 1}{( 2 t ) ^{4}} d t

y = ln (\frac{3 + e ^{x}}{3 - e ^{x}})

f (x) = (ln (x))^{3}, at x = 9

What is the tangent of 1/(sqrt(10x)),\at x=9 ?
The tangent of 1/(sqrt(10x)),\at x=9 is y=-1/(54sqrt(10))x+1/(2sqrt(10))