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