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Question Number 10712 by Saham last updated on 23/Feb/17
If y = x^x ,  find  (dy/dx)
$$\mathrm{If}\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{x}} ,\:\:\mathrm{find}\:\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$
Answered by mrW1 last updated on 23/Feb/17
y=x^x   ln y=xln x  (1/y)×(dy/dx)=ln x+1  ⇒(dy/dx)=y(ln x+1)=x^x (1+ln x)=x^x +x^x ln x
$${y}={x}^{{x}} \\ $$$$\mathrm{ln}\:{y}={x}\mathrm{ln}\:{x} \\ $$$$\frac{\mathrm{1}}{{y}}×\frac{{dy}}{{dx}}=\mathrm{ln}\:{x}+\mathrm{1} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={y}\left(\mathrm{ln}\:{x}+\mathrm{1}\right)={x}^{{x}} \left(\mathrm{1}+\mathrm{ln}\:{x}\right)={x}^{{x}} +{x}^{{x}} \mathrm{ln}\:{x} \\ $$
Commented by Saham last updated on 23/Feb/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by mrW1 last updated on 23/Feb/17
an other way:  y=x^x =e^(xln x)   (dy/dx)=e^(xln x) ×(ln x+1)=x^x (1+ln x)
$${an}\:{other}\:{way}: \\ $$$${y}={x}^{{x}} ={e}^{{x}\mathrm{ln}\:{x}} \\ $$$$\frac{{dy}}{{dx}}={e}^{{x}\mathrm{ln}\:{x}} ×\left(\mathrm{ln}\:{x}+\mathrm{1}\right)={x}^{{x}} \left(\mathrm{1}+\mathrm{ln}\:{x}\right) \\ $$
Commented by Saham last updated on 23/Feb/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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