What is the tangent of f(x)= 1/(sqrt(2x)),\at x=9 ?

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

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tangent of f(x)= 1/(sqrt(2x)),\at x=9

tangent of

f (x) = \frac{1}{2 x}, at x = 9

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

Solution steps

Find the tangent point:

(9, \frac{1}{3 2})

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{1}{54 2}

Find the line with slope m=

- \frac{1}{54 2}

and passing through

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

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

f (x) = (x^{4} - \frac{5}{x ^{3}})^{- 4}

\int_{0}^{3} \frac{x}{x + 1} d x

n = 1 \sum \infty \frac{2 n}{3 n - 2}

\frac{d y}{d x} = (y - 1) x

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

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