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Question Number 83413 by jagoll last updated on 02/Mar/20

y = x ∣x∣  find y ′ ?

$$\mathrm{y}\:=\:\mathrm{x}\:\mid\mathrm{x}\mid \\ $$$$\mathrm{find}\:\mathrm{y}\:'\:? \\ $$

Commented by mr W last updated on 02/Mar/20

∣x∣=sign(x)x  y=sign(x)x^2   y′=sign(x)2x+x^2 (sign(x))′  =2sign(x)x  =2∣x∣

$$\mid{x}\mid={sign}\left({x}\right){x} \\ $$$${y}={sign}\left({x}\right){x}^{\mathrm{2}} \\ $$$${y}'={sign}\left({x}\right)\mathrm{2}{x}+{x}^{\mathrm{2}} \left({sign}\left({x}\right)\right)' \\ $$$$=\mathrm{2}{sign}\left({x}\right){x} \\ $$$$=\mathrm{2}\mid{x}\mid \\ $$

Commented by Kunal12588 last updated on 02/Mar/20

what is sign(x) ? signum fuction?

$${what}\:{is}\:{sign}\left({x}\right)\:?\:{signum}\:{fuction}? \\ $$

Commented by mr W last updated on 02/Mar/20

sign(x)= { ((1, if x>0)),((0, if x=0)),((−1, if x<0)) :}

$${sign}\left({x}\right)=\begin{cases}{\mathrm{1},\:{if}\:{x}>\mathrm{0}}\\{\mathrm{0},\:{if}\:{x}=\mathrm{0}}\\{−\mathrm{1},\:{if}\:{x}<\mathrm{0}}\end{cases} \\ $$

Answered by MJS last updated on 02/Mar/20

y= { ((−x^2 ; x<0)),((x^2 ; x≥0)) :}  y′= { ((−2x; x<0)),((2x; x≥0)) :}  ⇒ y′=2∣x∣

$${y}=\begin{cases}{−{x}^{\mathrm{2}} ;\:{x}<\mathrm{0}}\\{{x}^{\mathrm{2}} ;\:{x}\geqslant\mathrm{0}}\end{cases} \\ $$$${y}'=\begin{cases}{−\mathrm{2}{x};\:{x}<\mathrm{0}}\\{\mathrm{2}{x};\:{x}\geqslant\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\:{y}'=\mathrm{2}\mid{x}\mid \\ $$

Commented by jagoll last updated on 02/Mar/20

yes...my answer right. thank you mister

$$\mathrm{yes}...\mathrm{my}\:\mathrm{answer}\:\mathrm{right}.\:\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$

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