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f-x-d-dx-f-x-




Question Number 131206 by Study last updated on 02/Feb/21
f(x)=∞  (d/dx)f(x)=?
$${f}\left({x}\right)=\infty \\ $$$$\frac{{d}}{{dx}}{f}\left({x}\right)=? \\ $$
Commented by mr W last updated on 02/Feb/21
∞ is not a variable, is not a constant.  f(x)=∞ has no meaning!  lim_(x→a) f(x)=∞ has meaning.
$$\infty\:{is}\:{not}\:{a}\:{variable},\:{is}\:{not}\:{a}\:{constant}. \\ $$$${f}\left({x}\right)=\infty\:{has}\:{no}\:{meaning}! \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)=\infty\:{has}\:{meaning}. \\ $$
Commented by JDamian last updated on 02/Feb/21
∞ is not a real number  ∞ ∉ ℜ
$$\infty\:{is}\:{not}\:{a}\:{real}\:{number} \\ $$$$\infty\:\notin\:\Re \\ $$
Commented by mr W last updated on 02/Feb/21
∞ is not a number at all!
$$\infty\:{is}\:{not}\:{a}\:{number}\:{at}\:{all}! \\ $$
Commented by MJS_new last updated on 03/Feb/21
I still believe that if f(x)=[term without x]  ⇒ (d/dx)[f(x)]=0  example:  f(x)=(1/r); r∈R ⇒ (d/dx)[f(x)]=0  we have the limit  f′(p)=lim_(h→0)  ((f(p+h)−f(p−h))/(2h))  in this case  f′(p)=lim_(h→0)  lim_(r→0)  (((1/r)−(1/r))/(2h)) =  =lim_(h→0)  lim_(r→0)  (0/(2h)) =0  if I′m wrong please prove
$$\mathrm{I}\:\mathrm{still}\:\mathrm{believe}\:\mathrm{that}\:\mathrm{if}\:{f}\left({x}\right)=\left[{term}\:{without}\:{x}\right] \\ $$$$\Rightarrow\:\frac{{d}}{{dx}}\left[{f}\left({x}\right)\right]=\mathrm{0} \\ $$$$\mathrm{example}: \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{r}};\:{r}\in\mathbb{R}\:\Rightarrow\:\frac{{d}}{{dx}}\left[{f}\left({x}\right)\right]=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{the}\:\mathrm{limit} \\ $$$${f}'\left({p}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({p}+{h}\right)−{f}\left({p}−{h}\right)}{\mathrm{2}{h}} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$${f}'\left({p}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{r}}}{\mathrm{2}{h}}\:= \\ $$$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{0}}{\mathrm{2}{h}}\:=\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{I}'\mathrm{m}\:\mathrm{wrong}\:\mathrm{please}\:\mathrm{prove} \\ $$
Commented by mr W last updated on 03/Feb/21
yes,f(x) musn′t have term with x, but  it must be validly defined and   represent values.
$${yes},{f}\left({x}\right)\:{musn}'{t}\:{have}\:{term}\:{with}\:{x},\:{but} \\ $$$${it}\:{must}\:{be}\:{validly}\:{defined}\:{and}\: \\ $$$${represent}\:{values}. \\ $$
Answered by prakash jain last updated on 03/Feb/21
(d/dx)(∞)=0
$$\frac{{d}}{{dx}}\left(\infty\right)=\mathrm{0} \\ $$