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

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

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tangent of sqrt(x)-1

tangent of

x - 1

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

Find the slope of

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

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

\frac{\partial}{\partial y} (x^{2} \cdot sin (2 y))

f (x) = x - 3 x

f (x) = x^{2} + x - 9

x \to \frac{1}{8} lim (\frac{1}{x} + x^{- 1})

\int \frac{1}{t ^{2} + 9} d t

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