What is the tangent of f(x)= 1/(2sqrt(x)),\at x=65 ?

The tangent of f(x)= 1/(2sqrt(x)),\at x=65 is y=-1/(260sqrt(65))x+3/(4sqrt(65))

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tangent of f(x)= 1/(2sqrt(x)),\at x=65

tangent of

f (x) = \frac{1}{2 x}, at x = 65

y = - \frac{1}{260 65} x + \frac{3}{4 65}

Solution steps

Find the tangent point:

(65, \frac{1}{2 65})

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 = 65 : m = - \frac{1}{260 65}

Find the line with slope m=

- \frac{1}{260 65}

and passing through

(65, \frac{1}{2 65}) : y = - \frac{1}{260 65} x + \frac{3}{4 65}

y = - \frac{1}{260 65} x + \frac{3}{4 65}

\frac{d}{d x} (cos (ln (x)))

\int_{0}^{1} \frac{10 θ}{( 1 + θ ^{\frac{3}{2}} ) ^{2}} d θ

\frac{d}{d x} (\frac{x}{( x ^{2} - 1 ) ^{2}})

f (x) = \frac{e ^{x}}{x ^{6}}

\int 4 t^{3} t^{2} + 1 d t

What is the tangent of f(x)= 1/(2sqrt(x)),\at x=65 ?
The tangent of f(x)= 1/(2sqrt(x)),\at x=65 is y=-1/(260sqrt(65))x+3/(4sqrt(65))