Question-82303

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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? (d/dx)f(x)=lim_(h→0) ((f(x+h)−f(x))/h)

$${this}\:{is}\:{the}\:{definition}\:{of}\:{derivative} \\ $$$${isn}'{t}\:{it}? \\ $$$$\frac{{d}}{{dx}}{f}\left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$

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 (x+h)^(x+h)

$${I}\:{think}\:{it}\:{is}\:{wrong}\:{because}\:{it}\:{should}\:{be} \\ $$$$\left({x}+{h}\right)^{{x}+{h}} \\ $$

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 (d/dx)(ln y) = lim_(h→0) (((x+h) ln(x+h)−xln x)/h) = lim_(h→0) ((x ( ln (1+(h/x)))−xln x)/h) = lim_(h→0) ((ln(1+(h/x))−ln x)/(((h/x))))

$${ln}\:{y}\:=\:{x}\:{ln}\:{x} \\ $$$$\frac{{d}}{{dx}}\left({ln}\:{y}\right)\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({x}+{h}\right)\:{ln}\left({x}+{h}\right)−{xln}\:{x}}{{h}} \\ $$$$=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\left(\:{ln}\:\left(\mathrm{1}+\frac{{h}}{{x}}\right)\right)−{xln}\:{x}}{{h}} \\ $$$$=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{ln}\left(\mathrm{1}+\frac{{h}}{{x}}\right)−{ln}\:{x}}{\left(\frac{{h}}{{x}}\right)} \\ $$

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

lim_(h→0) (((x+h)^(x+h) −x^x )/h) lim_(h→0) ((x^x [x^h (1+(h/x))^(x+h) −1])/h) x^x lim_(h→0) (((1+(h/x))^(x+h) [(x^h −1)]+(1+(h/x))^(x+h) −1)/h) x^x lim_(h→0) (1+(h/x))^(x+h) ×[(((x^h −1))/h)]+x^x lim_(h→0) (((1+(h/x))^(x+h) −1)/h) =x^x ×1×lnx+x^x lim_(h→0) (((1+(h/x))^(x+h) −(1)^(x+h) )/h) for second limit... k=1+(h/x) when h→0 k→1 x^x lim_(k→1) ((k^(x+h) −(1)^(x+h) )/(k−1))×(1/x) lim_(h→0) x^(x−1) ×(x+h)^ (1)^(x+h−1) =x^x so answer is x^x lnx+x^x =x^x (1+lnx) pls check.... Δ =li_(h→0) by direct method y=x^x →lny=xlnx (1/y)(dy/dx)=x×(1/x)+lnx→x^x (1+lnx)

$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({x}+{h}\right)^{{x}+{h}} −{x}^{{x}} }{{h}} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{{x}} \left[{x}^{{h}} \left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} −\mathrm{1}\right]}{{h}} \\ $$$${x}^{{x}} \underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} \left[\left({x}^{{h}} −\mathrm{1}\right)\right]+\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} −\mathrm{1}}{{h}} \\ $$$${x}^{{x}} \underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} ×\left[\frac{\left({x}^{{h}} −\mathrm{1}\right)}{{h}}\right]+\boldsymbol{{x}}^{\boldsymbol{{x}}} \underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{\boldsymbol{{h}}}{\boldsymbol{{x}}}\right)^{\boldsymbol{{x}}+\boldsymbol{{h}}} −\mathrm{1}}{\boldsymbol{{h}}} \\ $$$$=\boldsymbol{{x}}^{\boldsymbol{{x}}} ×\mathrm{1}×\boldsymbol{{lnx}}+\boldsymbol{{x}}^{\boldsymbol{{x}}} \underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{\boldsymbol{{h}}}{\boldsymbol{{x}}}\right)^{\boldsymbol{{x}}+\boldsymbol{{h}}} −\left(\mathrm{1}\right)^{\boldsymbol{{x}}+\boldsymbol{{h}}} }{\boldsymbol{{h}}} \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{second}}\:\boldsymbol{{limit}}… \\ $$$$\boldsymbol{{k}}=\mathrm{1}+\frac{\boldsymbol{{h}}}{\boldsymbol{{x}}}\:\:\boldsymbol{{when}}\:\boldsymbol{{h}}\rightarrow\mathrm{0}\:\:\:\:\:{k}\rightarrow\mathrm{1} \\ $$$${x}^{{x}} \underset{{k}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{k}^{{x}+{h}} −\left(\mathrm{1}\right)^{{x}+{h}} }{\mathrm{k}−\mathrm{1}}×\frac{\mathrm{1}}{{x}} \\ $$$$\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{{x}−\mathrm{1}} ×\left({x}+{h}\right)^{} \left(\mathrm{1}\right)^{{x}+{h}−\mathrm{1}} \\ $$$$=\boldsymbol{{x}}^{\boldsymbol{{x}}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}}\:\boldsymbol{{x}}^{\boldsymbol{{x}}} \boldsymbol{{lnx}}+\boldsymbol{{x}}^{\boldsymbol{{x}}} \\ $$$$=\boldsymbol{{x}}^{\boldsymbol{{x}}} \left(\mathrm{1}+\boldsymbol{{lnx}}\right) \\ $$$${pls}\:{check}…. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\Delta \\ $$$$ \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{li}} \\ $$$$\boldsymbol{{by}}\:\boldsymbol{{direct}}\:\boldsymbol{{method}} \\ $$$${y}={x}^{{x}} \rightarrow{lny}={xlnx} \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}={x}×\frac{\mathrm{1}}{{x}}+{lnx}\rightarrow\boldsymbol{{x}}^{\boldsymbol{{x}}} \left(\mathrm{1}+\boldsymbol{{lnx}}\right) \\ $$$$ \\ $$

Commented by Power last updated on 20/Feb/20

Commented by TANMAY PANACEA last updated on 20/Feb/20

lil_(h→0) look pls i think it is correct.. lim_(h→0) ((x^h −1)/h)=lnx l Δ

$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{li}{l}look}\:{pls} \\ $$$${i}\:{think}\:{it}\:{is}\:{correct}.. \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{{h}} −\mathrm{1}}{{h}}={lnx} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${l} \\ $$$$\Delta \\ $$$$ \\ $$

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=(1+(h/x))^(x+h) taking out x^x so x^h (1+(h/x))^(x+h) −1 =A(B−1) =B(A−1)+B−1 =(1+(h/x))^(x+h) (x^h −1)+(1+(h/x))^(x+h) −1

$${Let}\:{A}={x}^{{h}} \:\:\:\:{B}=\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} \\ $$$${taking}\:{out}\:{x}^{{x}} \\ $$$${so}\:{x}^{{h}} \left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} −\mathrm{1} \\ $$$$={A}\left({B}−\mathrm{1}\right) \\ $$$$={B}\left({A}−\mathrm{1}\right)+{B}−\mathrm{1} \\ $$$$=\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} \left({x}^{{h}} −\mathrm{1}\right)+\left(\mathrm{1}+\frac{{h}}{{x}}\right)^{{x}+{h}} −\mathrm{1} \\ $$

Commented by mathmax by abdo last updated on 20/Feb/20

(d/dx)(x^x ) =lim_(t→x) ((t^t −x^x )/(t−x)) changement t−x=h give (d/dx)(x^x ) =lim_(h→0) (((x+h)^(x+h) −x^x )/h) but we have (((x+h)^(x+h) −x^x )/h) =((e^((x+h)ln(x+h)) −e^(xln(x)) )/h) let f(h)=ln(x+h) ⇒f(x)=Σ_(n=0) ^∞ (h^n /(n!))f^((n)) (0) =f(o)+hf^′ (0)+ξ(h) =ln(x)+(h/x) +ξ(h^2 ) ⇒ (x+h)ln(x+h) =(x+h){lnx+(h/x) +ξ(h^2 )} =xlnx+h +xξ(h)+hln(x)+(h^2 /x) +hξ(h) ⇒ e^((x+h)ln(x+h)) =e^(xlnx +h(1+lnx) +(h^2 /x) +(x+h)ξ(h)) ⇒ (((x+h)^(x+h) −x^x )/h) =e^(xlnx) ×((e^(h(1+lnx)+(h^2 /x)+(x+h)ξ(h^2 )) −1)/h) ∼e^(xlnx) {((1+h(1+lnx)+(h^2 /x)+(x+h)ξ(h^2 )−1)/h)} =e^(xlnx) {1+lnx +(h/x) +(x+h)ξ(h)}→e^(xlnx) (1+lnx) (h→0) ⇒ (d/dx)(x^x ) =(1+lnx)x^x

$$\frac{{d}}{{dx}}\left({x}^{{x}} \right)\:={lim}_{{t}\rightarrow{x}} \:\:\:\frac{{t}^{{t}} −{x}^{{x}} }{{t}−{x}}\:\:\:{changement}\:{t}−{x}={h}\:{give} \\ $$$$\frac{{d}}{{dx}}\left({x}^{{x}} \right)\:={lim}_{{h}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{\left({x}+{h}\right)^{{x}+{h}} −{x}^{{x}} }{{h}}\:\:{but}\:{we}\:{have} \\ $$$$\frac{\left({x}+{h}\right)^{{x}+{h}} −{x}^{{x}} }{{h}}\:=\frac{{e}^{\left({x}+{h}\right){ln}\left({x}+{h}\right)} −{e}^{{xln}\left({x}\right)} }{{h}} \\ $$$${let}\:{f}\left({h}\right)={ln}\left({x}+{h}\right)\:\Rightarrow{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{h}^{{n}} }{{n}!}{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$={f}\left({o}\right)+{hf}^{'} \left(\mathrm{0}\right)+\xi\left({h}\right)\:={ln}\left({x}\right)+\frac{{h}}{{x}}\:+\xi\left({h}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\left({x}+{h}\right){ln}\left({x}+{h}\right)\:=\left({x}+{h}\right)\left\{{lnx}+\frac{{h}}{{x}}\:+\xi\left({h}^{\mathrm{2}} \right)\right\} \\ $$$$={xlnx}+{h}\:+{x}\xi\left({h}\right)+{hln}\left({x}\right)+\frac{{h}^{\mathrm{2}} }{{x}}\:+{h}\xi\left({h}\right)\:\Rightarrow \\ $$$${e}^{\left({x}+{h}\right){ln}\left({x}+{h}\right)} \:={e}^{{xlnx}\:+{h}\left(\mathrm{1}+{lnx}\right)\:+\frac{{h}^{\mathrm{2}} }{{x}}\:+\left({x}+{h}\right)\xi\left({h}\right)} \:\Rightarrow \\ $$$$\frac{\left({x}+{h}\right)^{{x}+{h}} −{x}^{{x}} }{{h}}\:={e}^{{xlnx}} ×\frac{{e}^{{h}\left(\mathrm{1}+{lnx}\right)+\frac{{h}^{\mathrm{2}} }{{x}}+\left({x}+{h}\right)\xi\left({h}^{\mathrm{2}} \right)} −\mathrm{1}}{{h}} \\ $$$$\sim{e}^{{xlnx}} \left\{\frac{\mathrm{1}+{h}\left(\mathrm{1}+{lnx}\right)+\frac{{h}^{\mathrm{2}} }{{x}}+\left({x}+{h}\right)\xi\left({h}^{\mathrm{2}} \right)−\mathrm{1}}{{h}}\right\} \\ $$$$={e}^{{xlnx}} \left\{\mathrm{1}+{lnx}\:+\frac{{h}}{{x}}\:+\left({x}+{h}\right)\xi\left({h}\right)\right\}\rightarrow{e}^{{xlnx}} \left(\mathrm{1}+{lnx}\right)\:\left({h}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\frac{{d}}{{dx}}\left({x}^{{x}} \right)\:=\left(\mathrm{1}+{lnx}\right){x}^{{x}} \: \\ $$

Commented by mathmax by abdo last updated on 20/Feb/20