What is the tangent of f(x)= 1/(2sqrt(x)),\at x=25 ?

The tangent of f(x)= 1/(2sqrt(x)),\at x=25 is y=-1/500 x+3/20

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tangent of f(x)= 1/(2sqrt(x)),\at x=25

tangent of

f (x) = \frac{1}{2 x}, at x = 25

y = - \frac{1}{500} x + \frac{3}{20}

Solution steps

Find the tangent point:

(25, \frac{1}{10})

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 = 25 : m = - \frac{1}{500}

Find the line with slope m=

- \frac{1}{500}

and passing through

(25, \frac{1}{10}) : y = - \frac{1}{500} x + \frac{3}{20}

y = - \frac{1}{500} x + \frac{3}{20}

- \frac{6 x}{x ^{2} + 1}

\frac{d}{d x} (ln ((arcsin (x))^{2}))

\frac{140000}{x}

\int \frac{1}{x ^{2} - 3 x - 15} d x

\int 40 + 6 x - x^{2} d x

What is the tangent of f(x)= 1/(2sqrt(x)),\at x=25 ?
The tangent of f(x)= 1/(2sqrt(x)),\at x=25 is y=-1/500 x+3/20