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

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

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tangent of sqrt(x+1)

tangent of

x + 1

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} + 1)

Find the slope of

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

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} + 1}

Find the line with slope m=

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

and passing through

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

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

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