What is the tangent of 2sqrt(x) ?

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

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tangent of 2sqrt(x)

tangent of

2 x

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0})

Find the slope of

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

Compute the slope of

y = 2 x

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

Find the line with slope m=

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

and passing through

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

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

\frac{x ^{4} + 36 x ^{2}}{( x ^{2} + 12 ) ^{2}}

x \to 2 lim (e^{x^{2} - 2 x})

\int \frac{x}{5 x - 4} d x

\int x (a^{2} - x^{2}) d x

f (x) = \frac{1}{2} x^{2}

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