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

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

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

tangent of

y = \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

y = \frac{1}{x} : \frac{d y}{d x} = - \frac{1}{2 x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{1}{x}

x = a_{0} : 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}}

\int_{1}^{\infty} x^{- 3} d x

y^{^{'}} - 5 y = 4 e^{7 t}

x \to 0 lim (x^{2} + a x + 2)

2 t^{2} - 3 t - 1

\int \frac{1}{4 x ^{2} - 81} d x

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