Question Number 82303 by Power last updated on 20/Feb/20
Commented by Power last updated on 20/Feb/20
$$\mathrm{thanks}\: \\ $$
Commented by Power last updated on 20/Feb/20
$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{it}} \\ $$
Commented by Power last updated on 20/Feb/20
$$\boldsymbol{\mathrm{derivative}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{limit}}\:\boldsymbol{\mathrm{from}}\:\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \\ $$
Commented by Tony Lin last updated on 20/Feb/20
$${this}\:{is}\:{the}\:{definition}\:{of}\:{derivative} \\ $$$${isn}'{t}\:{it}? \\ $$$$\frac{{d}}{{dx}}{f}\left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$
Commented by Power last updated on 20/Feb/20
$$\mathrm{yes}\:\mathrm{x}^{\mathrm{x}} \:\:\mathrm{must}\:\mathrm{be}\:\mathrm{proved}\:\mathrm{using}\:\mathrm{the}\:\mathrm{limit} \\ $$
Commented by JDamian last updated on 20/Feb/20
$${I}\:{think}\:{it}\:{is}\:{wrong}\:{because}\:{it}\:{should}\:{be} \\ $$$$\left({x}+{h}\right)^{{x}+{h}} \\ $$
Commented by Power last updated on 20/Feb/20
$$\mathrm{yes}\:\mathrm{sir} \\ $$
Commented by jagoll last updated on 20/Feb/20
$${ln}\:{y}\:=\:{x}\:{ln}\:{x} \\ $$$$\frac{{d}}{{dx}}\left({ln}\:{y}\right)\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({x}+{h}\right)\:{ln}\left({x}+{h}\right)−{xln}\:{x}}{{h}} \\ $$$$=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\left(\:{ln}\:\left(\mathrm{1}+\frac{{h}}{{x}}\right)\right)−{xln}\:{x}}{{h}} \\ $$$$=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{ln}\left(\mathrm{1}+\frac{{h}}{{x}}\right)−{ln}\:{x}}{\left(\frac{{h}}{{x}}\right)} \\ $$
Commented by Power last updated on 20/Feb/20
$$\mathrm{sir}\:\:\mathrm{prove}\:\mathrm{it} \\ $$
Commented by TANMAY PANACEA last updated on 20/Feb/20
$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({x}+{h}\right)^{{x}+{h}} −{x}^{{x}} }{{h}} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{{x}} \left[{x}^{{h}} \left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} −\mathrm{1}\right]}{{h}} \\ $$$${x}^{{x}} \underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} \left[\left({x}^{{h}} −\mathrm{1}\right)\right]+\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} −\mathrm{1}}{{h}} \\ $$$${x}^{{x}} \underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} ×\left[\frac{\left({x}^{{h}} −\mathrm{1}\right)}{{h}}\right]+\boldsymbol{{x}}^{\boldsymbol{{x}}} \underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{\boldsymbol{{h}}}{\boldsymbol{{x}}}\right)^{\boldsymbol{{x}}+\boldsymbol{{h}}} −\mathrm{1}}{\boldsymbol{{h}}} \\ $$$$=\boldsymbol{{x}}^{\boldsymbol{{x}}} ×\mathrm{1}×\boldsymbol{{lnx}}+\boldsymbol{{x}}^{\boldsymbol{{x}}} \underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{\boldsymbol{{h}}}{\boldsymbol{{x}}}\right)^{\boldsymbol{{x}}+\boldsymbol{{h}}} −\left(\mathrm{1}\right)^{\boldsymbol{{x}}+\boldsymbol{{h}}} }{\boldsymbol{{h}}} \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{second}}\:\boldsymbol{{limit}}… \\ $$$$\boldsymbol{{k}}=\mathrm{1}+\frac{\boldsymbol{{h}}}{\boldsymbol{{x}}}\:\:\boldsymbol{{when}}\:\boldsymbol{{h}}\rightarrow\mathrm{0}\:\:\:\:\:{k}\rightarrow\mathrm{1} \\ $$$${x}^{{x}} \underset{{k}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{k}^{{x}+{h}} −\left(\mathrm{1}\right)^{{x}+{h}} }{\mathrm{k}−\mathrm{1}}×\frac{\mathrm{1}}{{x}} \\ $$$$\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{{x}−\mathrm{1}} ×\left({x}+{h}\right)^{} \left(\mathrm{1}\right)^{{x}+{h}−\mathrm{1}} \\ $$$$=\boldsymbol{{x}}^{\boldsymbol{{x}}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}}\:\boldsymbol{{x}}^{\boldsymbol{{x}}} \boldsymbol{{lnx}}+\boldsymbol{{x}}^{\boldsymbol{{x}}} \\ $$$$=\boldsymbol{{x}}^{\boldsymbol{{x}}} \left(\mathrm{1}+\boldsymbol{{lnx}}\right) \\ $$$${pls}\:{check}…. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\Delta \\ $$$$ \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{li}} \\ $$$$\boldsymbol{{by}}\:\boldsymbol{{direct}}\:\boldsymbol{{method}} \\ $$$${y}={x}^{{x}} \rightarrow{lny}={xlnx} \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}={x}×\frac{\mathrm{1}}{{x}}+{lnx}\rightarrow\boldsymbol{{x}}^{\boldsymbol{{x}}} \left(\mathrm{1}+\boldsymbol{{lnx}}\right) \\ $$$$ \\ $$
Commented by Power last updated on 20/Feb/20
Commented by TANMAY PANACEA last updated on 20/Feb/20
$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{li}{l}look}\:{pls} \\ $$$${i}\:{think}\:{it}\:{is}\:{correct}.. \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{{h}} −\mathrm{1}}{{h}}={lnx} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${l} \\ $$$$\Delta \\ $$$$ \\ $$
Commented by TANMAY PANACEA last updated on 20/Feb/20
$${pls}\:{check}…{i}\:{have}\:{solved} \\ $$
Commented by TANMAY PANACEA last updated on 20/Feb/20
$${Let}\:{A}={x}^{{h}} \:\:\:\:{B}=\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} \\ $$$${taking}\:{out}\:{x}^{{x}} \\ $$$${so}\:{x}^{{h}} \left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} −\mathrm{1} \\ $$$$={A}\left({B}−\mathrm{1}\right) \\ $$$$={B}\left({A}−\mathrm{1}\right)+{B}−\mathrm{1} \\ $$$$=\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} \left({x}^{{h}} −\mathrm{1}\right)+\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} −\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 20/Feb/20
$$\frac{{d}}{{dx}}\left({x}^{{x}} \right)\:={lim}_{{t}\rightarrow{x}} \:\:\:\frac{{t}^{{t}} −{x}^{{x}} }{{t}−{x}}\:\:\:{changement}\:{t}−{x}={h}\:{give} \\ $$$$\frac{{d}}{{dx}}\left({x}^{{x}} \right)\:={lim}_{{h}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{\left({x}+{h}\right)^{{x}+{h}} −{x}^{{x}} }{{h}}\:\:{but}\:{we}\:{have} \\ $$$$\frac{\left({x}+{h}\right)^{{x}+{h}} −{x}^{{x}} }{{h}}\:=\frac{{e}^{\left({x}+{h}\right){ln}\left({x}+{h}\right)} −{e}^{{xln}\left({x}\right)} }{{h}} \\ $$$${let}\:{f}\left({h}\right)={ln}\left({x}+{h}\right)\:\Rightarrow{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{h}^{{n}} }{{n}!}{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$={f}\left({o}\right)+{hf}^{'} \left(\mathrm{0}\right)+\xi\left({h}\right)\:={ln}\left({x}\right)+\frac{{h}}{{x}}\:+\xi\left({h}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\left({x}+{h}\right){ln}\left({x}+{h}\right)\:=\left({x}+{h}\right)\left\{{lnx}+\frac{{h}}{{x}}\:+\xi\left({h}^{\mathrm{2}} \right)\right\} \\ $$$$={xlnx}+{h}\:+{x}\xi\left({h}\right)+{hln}\left({x}\right)+\frac{{h}^{\mathrm{2}} }{{x}}\:+{h}\xi\left({h}\right)\:\Rightarrow \\ $$$${e}^{\left({x}+{h}\right){ln}\left({x}+{h}\right)} \:={e}^{{xlnx}\:+{h}\left(\mathrm{1}+{lnx}\right)\:+\frac{{h}^{\mathrm{2}} }{{x}}\:+\left({x}+{h}\right)\xi\left({h}\right)} \:\Rightarrow \\ $$$$\frac{\left({x}+{h}\right)^{{x}+{h}} −{x}^{{x}} }{{h}}\:={e}^{{xlnx}} ×\frac{{e}^{{h}\left(\mathrm{1}+{lnx}\right)+\frac{{h}^{\mathrm{2}} }{{x}}+\left({x}+{h}\right)\xi\left({h}^{\mathrm{2}} \right)} −\mathrm{1}}{{h}} \\ $$$$\sim{e}^{{xlnx}} \left\{\frac{\mathrm{1}+{h}\left(\mathrm{1}+{lnx}\right)+\frac{{h}^{\mathrm{2}} }{{x}}+\left({x}+{h}\right)\xi\left({h}^{\mathrm{2}} \right)−\mathrm{1}}{{h}}\right\} \\ $$$$={e}^{{xlnx}} \left\{\mathrm{1}+{lnx}\:+\frac{{h}}{{x}}\:+\left({x}+{h}\right)\xi\left({h}\right)\right\}\rightarrow{e}^{{xlnx}} \left(\mathrm{1}+{lnx}\right)\:\left({h}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\frac{{d}}{{dx}}\left({x}^{{x}} \right)\:=\left(\mathrm{1}+{lnx}\right){x}^{{x}} \: \\ $$
Commented by mathmax by abdo last updated on 20/Feb/20
$${line}\:\mathrm{5}\:{f}\left({h}\right)={f}\left({o}+{hf}^{'} \left({o}\right)\:+{h}^{\mathrm{2}} \xi\left({h}\right)….\right. \\ $$
Commented by Power last updated on 20/Feb/20
$$\mathrm{thanks} \\ $$
Commented by mathmax by abdo last updated on 20/Feb/20
$${you}\:{are}\:{welcome}. \\ $$