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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=sqrt(x)

tangent of

y = x

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0})

Find the slope of

y = x : \frac{d y}{d x} = \frac{1}{2 x}

Compute the slope of

y = x

x = a_{0} : m = \frac{1}{2 a _{0}}

Find the line with slope m=

\frac{1}{2 a _{0}}

and passing through

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

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

y^{^{''}} - 2 y^{^{'}} + y = 14.5 e^{t}

\int (1 - y^{2})^{- \frac{1}{2}} d y

y^{^{'}} + \frac{y}{x} = 9 x^{3} y^{2}

\frac{d}{d x} (n^{x})

y^{^{''}} + 6 y^{^{'}} + 9 y = 0

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