What is the tangent of y=2sqrt(x) ?

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

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tangent of y=2sqrt(x)

tangent of

y = 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}

\int e^{x} e^{3 x} d x

5 x 4 - 5 x, at x = - 2

5 x^{2} \frac{d y}{d x} = \frac{d y}{d x} + 9 x e^{- y}

\frac{d}{d x} (4 sec^{2} (4 x))

5 e^{t}

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