What is the tangent of y=16sqrt(x) ?

The tangent of y=16sqrt(x) is y= 8/(sqrt(a_{0))}x+8sqrt(a_{0)}

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tangent of y=16sqrt(x)

tangent of

y = 16 x

y = \frac{8}{a _{0}} x + 8 a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 16 a_{0})

Find the slope of

y = 16 x : \frac{d y}{d x} = \frac{8}{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{8}{a _{0}}

Find the line with slope m=

\frac{8}{a _{0}}

and passing through

(a_{0}, 16 a_{0}) : y = \frac{8}{a _{0}} x + 8 a_{0}

y = \frac{8}{a _{0}} x + 8 a_{0}

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\int (\frac{1}{x ^{2}} - x^{2} - \frac{1}{3}) d x

\int \frac{2 x}{x ^{2} - 2 x - 3} d x

What is the tangent of y=16sqrt(x) ?
The tangent of y=16sqrt(x) is y= 8/(sqrt(a_{0))}x+8sqrt(a_{0)}