What is the tangent of y=4sqrt(x) ?

The tangent of y=4sqrt(x) is y= 2/(sqrt(a_{0))}x+2sqrt(a_{0)}

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tangent of y=4sqrt(x)

tangent of

y = 4 x

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0})

Find the slope of

y = 4 x : \frac{d y}{d x} = \frac{2}{x}

Compute the slope of

y = 4 x

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

Find the line with slope m=

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

and passing through

(a_{0}, 4 a_{0}) : y = \frac{2}{a _{0}} x + 2 a_{0}

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

\frac{\partial}{\partial x} (x^{3} y^{2} - 4 y x^{2} + 3 y x - 4)

f (x) = \frac{x}{x ^{2} - 9}

y = x^{2} + 5 x + 6, at x = 7

\int 144 - 9 x^{2} d x

x \to - 5 + lim (\frac{3}{x ^{2} - 25})

What is the tangent of y=4sqrt(x) ?
The tangent of y=4sqrt(x) is y= 2/(sqrt(a_{0))}x+2sqrt(a_{0)}