What is the tangent of x|x| ?

The tangent of x|x| is y=(2a_{0}^2)/(|a_{0)|}x+a_{0}|a_{0}|-(2a_{0}^3)/(|a_{0)|}

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tangent of x|x|

tangent of

x ∣ x ∣

y = \frac{2 a _{0}^{2}}{∣ a _{0} ∣} x + a_{0} ∣ a_{0} ∣ - \frac{2 a _{0}^{3}}{∣ a _{0} ∣}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} ∣ a_{0} ∣)

Find the slope of

y = x ∣ x ∣ : \frac{d y}{d x} = \frac{2 x ^{2}}{∣ 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{2 a _{0}^{2}}{∣ a _{0} ∣}

Find the line with slope m=

\frac{2 a _{0}^{2}}{∣ a _{0} ∣}

and passing through

(a_{0}, a_{0} ∣ a_{0} ∣) : y = \frac{2 a _{0}^{2}}{∣ a _{0} ∣} x + a_{0} ∣ a_{0} ∣ - \frac{2 a _{0}^{3}}{∣ a _{0} ∣}

y = \frac{2 a _{0}^{2}}{∣ a _{0} ∣} x + a_{0} ∣ a_{0} ∣ - \frac{2 a _{0}^{3}}{∣ a _{0} ∣}

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

x \to 2 + lim (\frac{x + 1}{x - 2})

\int_{0}^{π} 5 sin^{2} (t co) s^{4} t d t

f (x) = (ln (x))^{4}, at x = 2

y = \frac{x ^{3}}{5 - x ^{2}}

What is the tangent of x|x| ?
The tangent of x|x| is y=(2a_{0}^2)/(|a_{0)|}x+a_{0}|a_{0}|-(2a_{0}^3)/(|a_{0)|}