What is the tangent of f(x)=sqrt(x+3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=sqrt(x+3)

tangent of

f (x) = x + 3

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} + 3)

Find the slope of

f (x) = x + 3 : \frac{df ( x )}{d x} = \frac{1}{2 x + 3}

Compute the slope of

f (x) = x + 3

x = a_{0} : m = \frac{1}{2 a _{0} + 3}

Find the line with slope m=

\frac{1}{2 a _{0} + 3}

and passing through

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

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

\frac{d y}{d t} = (3 - 2 y) \cdot t^{3}

\frac{\partial}{\partial z} (- 5 z)

y^{^{'}} = 8 y

\frac{d}{d x} (ln (1 + \frac{4}{x}))

y^{^{'}} = (y^{2} - 1) x

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