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

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

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tangent of y=1+sqrt(x)

tangent of

y = 1 + x

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

Find the slope of

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

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}}

Find the line with slope m=

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

and passing through

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

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

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