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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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