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}

Compute the slope of

y = x - 1

x = a_{0} : 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}

4 x

x \to - 3 lim (x^{4} - 3 x + 10)

G (t) = \frac{1 - 4 t}{5 + t}

\int \frac{3}{8 + 3 x} d x

\int 5 - 4 x - 4 x^{2} d x

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)}