What is the tangent of f(x)= 1/(2sqrt(x)),\at x=36 ?

The tangent of f(x)= 1/(2sqrt(x)),\at x=36 is y=-1/864 x+1/8

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tangent of f(x)= 1/(2sqrt(x)),\at x=36

tangent of

f (x) = \frac{1}{2 x}, at x = 36

y = - \frac{1}{864} x + \frac{1}{8}

Solution steps

Find the tangent point:

(36, \frac{1}{12})

Find the slope of

f (x) = \frac{1}{2 x} : \frac{df ( x )}{d x} = - \frac{1}{4 x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{1}{2 x}

x = 36 : m = - \frac{1}{864}

Find the line with slope m=

- \frac{1}{864}

and passing through

(36, \frac{1}{12}) : y = - \frac{1}{864} x + \frac{1}{8}

y = - \frac{1}{864} x + \frac{1}{8}

f (x) = \frac{tan ( 9 x )}{1 + sin ( 9 x )}

(2 x + 3 y - 1) d x - 4 (x + 1) d y = 0

\frac{e ^{- 2 s}}{s ( s - 2 )}

\frac{d}{d x} (- 2 e^{5 y} x - 2 sin (2 x))

\int_{- 2}^{2} (4 - x^{2}) d x

What is the tangent of f(x)= 1/(2sqrt(x)),\at x=36 ?
The tangent of f(x)= 1/(2sqrt(x)),\at x=36 is y=-1/864 x+1/8