What is the tangent of f(x)= 1/(sqrt(x)) ?

The tangent of f(x)= 1/(sqrt(x)) is y=-1/(2a_{0)^{3/2}}x+3/(2sqrt(a_{0))}

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tangent of f(x)= 1/(sqrt(x))

tangent of

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

y = - \frac{1}{2 a _{0}^{\frac{3}{2}}} x + \frac{3}{2 a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{a _{0}})

Find the slope of

f (x) = \frac{1}{x} : \frac{df ( x )}{d x} = - \frac{1}{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}{2 a _{0}^{\frac{3}{2}}}

Find the line with slope m=

- \frac{1}{2 a _{0}^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{1}{a _{0}}) : y = - \frac{1}{2 a _{0}^{\frac{3}{2}}} x + \frac{3}{2 a _{0}}

y = - \frac{1}{2 a _{0}^{\frac{3}{2}}} x + \frac{3}{2 a _{0}}

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\frac{d}{d x} (ln (x + 9 + x^{2}))

What is the tangent of f(x)= 1/(sqrt(x)) ?
The tangent of f(x)= 1/(sqrt(x)) is y=-1/(2a_{0)^{3/2}}x+3/(2sqrt(a_{0))}