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Question-82303




Question Number 82303 by Power last updated on 20/Feb/20
Commented by Power last updated on 20/Feb/20
thanks
thanks
Commented by Power last updated on 20/Feb/20
prove it
proveit
Commented by Power last updated on 20/Feb/20
derivative of a limit from  x^x
derivativeofalimitfromxx
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)
thisisthedefinitionofderivativeisntit?ddxf(x)=limh0f(x+h)f(x)h
Commented by Power last updated on 20/Feb/20
yes x^x   must be proved using the limit
yesxxmustbeprovedusingthelimit
Commented by JDamian last updated on 20/Feb/20
I think it is wrong because it should be  (x+h)^(x+h)
Ithinkitiswrongbecauseitshouldbe(x+h)x+h
Commented by Power last updated on 20/Feb/20
yes sir
yessir
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))))
lny=xlnxddx(lny)=limh0(x+h)ln(x+h)xlnxh=limh0x(ln(1+hx))xlnxh=limh0ln(1+hx)lnx(hx)
Commented by Power last updated on 20/Feb/20
sir  prove it
sirproveit
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)
limh0(x+h)x+hxxhlimh0xx[xh(1+hx)x+h1]hxxlimh0(1+hx)x+h[(xh1)]+(1+hx)x+h1hxxlimh0(1+hx)x+h×[(xh1)h]+xxlimh0(1+hx)x+h1h=xx×1×lnx+xxlimh0(1+hx)x+h(1)x+hhforsecondlimitk=1+hxwhenh0k1xxlimk1kx+h(1)x+hk1×1xlimh0xx1×(x+h)(1)x+h1=xxsoanswerisxxlnx+xx=xx(1+lnx)plscheck.Δ=lih0bydirectmethody=xxlny=xlnx1ydydx=x×1x+lnxxx(1+lnx)
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
lillookh0plsithinkitiscorrect..limh0xh1h=lnxlΔ
Commented by TANMAY PANACEA last updated on 20/Feb/20
pls check...i have solved
plscheckihavesolved
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
LetA=xhB=(1+hx)x+htakingoutxxsoxh(1+hx)x+h1=A(B1)=B(A1)+B1=(1+hx)x+h(xh1)+(1+hx)x+h1
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
ddx(xx)=limtxttxxtxchangementtx=hgiveddx(xx)=limh0(x+h)x+hxxhbutwehave(x+h)x+hxxh=e(x+h)ln(x+h)exln(x)hletf(h)=ln(x+h)f(x)=n=0hnn!f(n)(0)=f(o)+hf(0)+ξ(h)=ln(x)+hx+ξ(h2)(x+h)ln(x+h)=(x+h){lnx+hx+ξ(h2)}=xlnx+h+xξ(h)+hln(x)+h2x+hξ(h)e(x+h)ln(x+h)=exlnx+h(1+lnx)+h2x+(x+h)ξ(h)(x+h)x+hxxh=exlnx×eh(1+lnx)+h2x+(x+h)ξ(h2)1hexlnx{1+h(1+lnx)+h2x+(x+h)ξ(h2)1h}=exlnx{1+lnx+hx+(x+h)ξ(h)}exlnx(1+lnx)(h0)ddx(xx)=(1+lnx)xx
Commented by mathmax by abdo last updated on 20/Feb/20
line 5 f(h)=f(o+hf^′ (o) +h^2 ξ(h)....
line5f(h)=f(o+hf(o)+h2ξ(h).
Commented by Power last updated on 20/Feb/20
thanks
thanks
Commented by mathmax by abdo last updated on 20/Feb/20
you are welcome.
youarewelcome.

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