What is the tangent of y=2sqrt(x)+1 ?

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

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tangent of y=2sqrt(x)+1

tangent of

y = 2 x + 1

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0} + 1)

Find the slope of

y = 2 x + 1 : \frac{d y}{d x} = \frac{1}{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}{a _{0}}

Find the line with slope m=

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

and passing through

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

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

y^{^{'}} + \frac{3}{x} y = 0, y (1) = 5

f (x) = e^{x} 1 + x^{2}

\frac{d}{d x} (ln (x^{2} + 2 - y))

\frac{\partial}{\partial x} (\frac{x}{sin ( 2 x ) + 3 y + z ^{2}})

\int sinh^{3} (x) d x

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