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

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

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

tangent of

f (x) = \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

f (x) = \frac{1}{10 x} : \frac{df ( x )}{d x} = - \frac{1}{2 10 x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \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}

\int x^{6} + \frac{1}{1 - 4 x ^{2}} d x

e^{- 4 t}

(i)^{^{'}}

\frac{d}{d x} (1 + tan^{2} (x))

\frac{d}{d x} (\frac{1}{( x ^{2} + y ^{2} )})

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