What is the tangent of sqrt(x) ?

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

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tangent of sqrt(x)

tangent of

x

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

Find the slope of

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

Compute the slope of

y = x

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

Find the line with slope m=

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

and passing through

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

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

2 x^{2}

y = 3 x + 2

y = 2 x + 3

(- 4.5) (0.8)

f (x) = sin^{3} (x)

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