let f(x) = ((∣x∣)/((1+∣1−x^2 ∣)^n )) study tbe derivability of f at points 0 and 1 (n natural integr)


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Question Number 40102 by maxmathsup by imad last updated on 15/Jul/18

$l e t f (x) = \frac{∣ x ∣}{{(1 + ∣ 1 - x^{2} ∣)}^{n}}$ $s t u d y t b e d e r i v a b i l i t y o f f a t p o i n t s 0 a n d 1 (n n a t u r a l i n t e g r)$
	Answered by math khazana by abdo last updated on 26/Jul/18

	$1) d e r i v a b i l i t y a t 0$ ${l i m}_{h \to 0^{+}} \frac{f (0 + h) - f (0)}{h} = {l i m}_{h \to 0^{+}} \frac{h}{(1 + {(1 - h^{2})}^{n}}$ $= {l i m}_{h \to 0^{+}} \frac{h}{h {(2 + h^{2})}^{n}} = {l i m}_{h \to 0^{+}} \frac{1}{{(2 + h^{2})}^{n}} = \frac{1}{2^{n}}$ ${l i m}_{h \to 0^{-}} \frac{f (0 + h) - f (0)}{h} = {l i m}_{h \to 0^{-}} \frac{- 1}{{(2 + h^{2})}^{n}} = \frac{- 1}{2^{n}}$ $w e c o n c l u d e t h a t f i s d e r i v a b l e a t l e f t a n d$ $r i g h t a t b u t n o t d e r i v a b l e a t t h i s p o i n t .$
	Answered by math khazana by abdo last updated on 28/Jul/18
	$2) derivability at 1 lim_(h→0^+ ) ((f(1+h) −f(1))/h) =lim_(h→0^+ ) ((∣1+h∣)/((1+∣1−(1+h)^2 ∣)^n )) −1 =lim_(h→0^+ ) ((1+h)/((1+((1+h)^2 −1))^n )) −1 =lim_(h→0^+ ) ((1+h)/((1+h^2 +2h)^n )) −1 =lim_(h→0^− ) ((1+h−(h+1)^(2n) )/((h+1)^(2n) )) =0 f is derivable at right of 1 lim_(h→0^− ) ((f(1+h)−f(1))/h)=lim_(h→0^− ) ((1+h)/({1+∣1−(1+h)^2 }^n )) −1 =lim_(h→0^− ) ((1+h)/({1+(1−(1+h)^2 }^n )) −1 =lim_(h→0^− ) ((1+h)/({1+(1−h^2 −2h−1)}^n )) −1 =lim_(h→0^− ) ((1+h)/({1−h^2 −2h)^n ))−1 =0 f is derivable at left of 1 we have f_g ^′ (1)=f_d ^′ (1) ⇒ f is derivable at x_0 =1$
	$2) d e r i v a b i l i t y a t 1$ ${l i m}_{h \to 0^{+}} \frac{f (1 + h) - f (1)}{h}$ $= {l i m}_{h \to 0^{+}} \frac{∣ 1 + h ∣}{{(1 + ∣ 1 - {(1 + h)}^{2} ∣)}^{n}} - 1$ $= {l i m}_{h \to 0^{+}} \frac{1 + h}{{(1 + ({(1 + h)}^{2} - 1))}^{n}} - 1$ $= {l i m}_{h \to 0^{+}} \frac{1 + h}{{(1 + h^{2} + 2 h)}^{n}} - 1$ $= {l i m}_{h \to 0^{-}} \frac{1 + h - {(h + 1)}^{2 n}}{{(h + 1)}^{2 n}} = 0 f i s d e r i v a b l e a t$ $r i g h t o f 1$ ${l i m}_{h \to 0^{-}} \frac{f (1 + h) - f (1)}{h} = {l i m}_{h \to 0^{-}} \frac{1 + h}{{1 + ∣ 1 - {(1 + h)}^{2}}^{n}} - 1$ $= {l i m}_{h \to 0^{-}} \frac{1 + h}{{1 + {(1 - {(1 + h)}^{2}}}^{n}} - 1$ $= {l i m}_{h \to 0^{-}} \frac{1 + h}{{1 + (1 - h^{2} - 2 h - 1)}^{n}} - 1$ $= {l i m}_{h \to 0^{-}} \frac{1 + h}{{{1 - h^{2} - 2 h)}^{n}} - 1 = 0$ $f i s d e r i v a b l e a t l e f t o f 1 w e h a v e$ $f_{g}^{^{'}} (1) = f_{d}^{^{'}} (1) \Rightarrow f i s d e r i v a b l e a t x_{0} = 1$
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