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y-x-x-dy-dx-




Question Number 77859 by jagoll last updated on 11/Jan/20
y = ∣x∣^(∣x∣)     (dy/dx) = ?
$${y}\:=\:\mid{x}\mid\:^{\mid{x}\mid} \:\: \\ $$$$\frac{{dy}}{{dx}}\:=\:? \\ $$
Commented by mathmax by abdo last updated on 11/Jan/20
y(x)=e^(∣x∣ln∣x∣)   if x>0 ⇒y(x)=e^(xln(x))  ⇒y^′ (x)=(xlnx)^′  e^(xlnx)   =(lnx+1)x^x   if x<0 ⇒y(x)=e^(−xln(−x))  ⇒y^′ (x)=(−xln(−x))^′  e^(−xln(−x))   =(−ln(−x)−x×((−1)/(−x)))(−x)^(−x) =(−x−ln(−x))(−x)^(−x)
$${y}\left({x}\right)={e}^{\mid{x}\mid{ln}\mid{x}\mid} \:\:{if}\:{x}>\mathrm{0}\:\Rightarrow{y}\left({x}\right)={e}^{{xln}\left({x}\right)} \:\Rightarrow{y}^{'} \left({x}\right)=\left({xlnx}\right)^{'} \:{e}^{{xlnx}} \\ $$$$=\left({lnx}+\mathrm{1}\right){x}^{{x}} \\ $$$${if}\:{x}<\mathrm{0}\:\Rightarrow{y}\left({x}\right)={e}^{−{xln}\left(−{x}\right)} \:\Rightarrow{y}^{'} \left({x}\right)=\left(−{xln}\left(−{x}\right)\right)^{'} \:{e}^{−{xln}\left(−{x}\right)} \\ $$$$=\left(−{ln}\left(−{x}\right)−{x}×\frac{−\mathrm{1}}{−{x}}\right)\left(−{x}\right)^{−{x}} =\left(−{x}−{ln}\left(−{x}\right)\right)\left(−{x}\right)^{−{x}} \\ $$
Answered by mr W last updated on 11/Jan/20
if x>0:  y=x^x   ln y=xln x  ((y′)/y)=ln x+1  y′=(1+ln x)x^x     if x<0:  y=(−x)^((−x))   ln y=−x ln (−x)  ((y′)/y)=−ln (−x)−1  y′=−(1+ln (−x))(−x)^((−x))     ⇒y′=sign(x)(1+ln ∣x∣)∣x∣^(∣x∣)
$${if}\:{x}>\mathrm{0}: \\ $$$${y}={x}^{{x}} \\ $$$$\mathrm{ln}\:{y}={x}\mathrm{ln}\:{x} \\ $$$$\frac{{y}'}{{y}}=\mathrm{ln}\:{x}+\mathrm{1} \\ $$$${y}'=\left(\mathrm{1}+\mathrm{ln}\:{x}\right){x}^{{x}} \\ $$$$ \\ $$$${if}\:{x}<\mathrm{0}: \\ $$$${y}=\left(−{x}\right)^{\left(−{x}\right)} \\ $$$$\mathrm{ln}\:{y}=−{x}\:\mathrm{ln}\:\left(−{x}\right) \\ $$$$\frac{{y}'}{{y}}=−\mathrm{ln}\:\left(−{x}\right)−\mathrm{1} \\ $$$${y}'=−\left(\mathrm{1}+\mathrm{ln}\:\left(−{x}\right)\right)\left(−{x}\right)^{\left(−{x}\right)} \\ $$$$ \\ $$$$\Rightarrow{y}'={sign}\left({x}\right)\left(\mathrm{1}+\mathrm{ln}\:\mid{x}\mid\right)\mid{x}\mid^{\mid{x}\mid} \\ $$
Commented by jagoll last updated on 11/Jan/20
what is sign (x) sir?
$${what}\:{is}\:{sign}\:\left({x}\right)\:{sir}? \\ $$
Commented by mr W last updated on 11/Jan/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 john santu last updated on 11/Jan/20
y = e^(∣x∣ ln(∣x∣))  ⇒ (dy/dx)= e^(∣x∣ ln(∣x∣))  ((x/(∣x∣)) ln(∣x∣)+ ∣x∣ (1/(∣x∣)) (x/(∣x∣)))  (dy/dx)= ∣x∣^(∣x∣)  ((x/(∣x∣))ln(∣x∣)+(x/(∣x∣)))
$${y}\:=\:{e}^{\mid{x}\mid\:{ln}\left(\mid{x}\mid\right)} \:\Rightarrow\:\frac{{dy}}{{dx}}=\:{e}^{\mid{x}\mid\:{ln}\left(\mid{x}\mid\right)} \:\left(\frac{{x}}{\mid{x}\mid}\:{ln}\left(\mid{x}\mid\right)+\:\mid{x}\mid\:\frac{\mathrm{1}}{\mid{x}\mid}\:\frac{{x}}{\mid{x}\mid}\right) \\ $$$$\frac{{dy}}{{dx}}=\:\mid{x}\mid^{\mid{x}\mid} \:\left(\frac{{x}}{\mid{x}\mid}{ln}\left(\mid{x}\mid\right)+\frac{{x}}{\mid{x}\mid}\right) \\ $$

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