What is the tangent of 8sqrt(x) ?

The tangent of 8sqrt(x) is y= 4/(sqrt(a_{0))}x+4sqrt(a_{0)}

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tangent of 8sqrt(x)

tangent of

8 x

y = \frac{4}{a _{0}} x + 4 a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0})

Find the slope of

y = 8 x : \frac{d y}{d x} = \frac{4}{x}

Compute the slope of

y = 8 x

x = a_{0} : m = \frac{4}{a _{0}}

Find the line with slope m=

\frac{4}{a _{0}}

and passing through

(a_{0}, 8 a_{0}) : y = \frac{4}{a _{0}} x + 4 a_{0}

y = \frac{4}{a _{0}} x + 4 a_{0}

\frac{m \cdot s}{s ^{2} + \frac{s}{n \cdot m}}

\frac{\partial}{\partial x} (y^{2} (2 x e^{2 x y} + e^{2 x y} \cdot 2 y x^{2}))

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

x \to 0 lim (\frac{4}{x ( x + 2 )})

(5 x - 3)^{4} (2 x - 7)^{- 3}

What is the tangent of 8sqrt(x) ?
The tangent of 8sqrt(x) is y= 4/(sqrt(a_{0))}x+4sqrt(a_{0)}