What is the tangent of 12sqrt(x) ?

The tangent of 12sqrt(x) is y= 6/(sqrt(a_{0))}x+6sqrt(a_{0)}

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tangent of 12sqrt(x)

tangent of

12 x

y = \frac{6}{a _{0}} x + 6 a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 12 a_{0})

Find the slope of

y = 12 x : \frac{d y}{d x} = \frac{6}{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{6}{a _{0}}

Find the line with slope m=

\frac{6}{a _{0}}

and passing through

(a_{0}, 12 a_{0}) : y = \frac{6}{a _{0}} x + 6 a_{0}

y = \frac{6}{a _{0}} x + 6 a_{0}

\int (5 x + 2)^{- 3} d x

\frac{d}{d x} (3 π x)

\int tan (4 x + 1) d x

\int (3 t - 1)^{3} d t

\frac{d y}{d t} + 2 y = 3

What is the tangent of 12sqrt(x) ?
The tangent of 12sqrt(x) is y= 6/(sqrt(a_{0))}x+6sqrt(a_{0)}