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Question Number 78897 by Mikael_786 last updated on 21/Jan/20

Commented by mind is power last updated on 21/Jan/20

$t r a n s l a t e p l e a s e$
Commented by mr W last updated on 21/Jan/20

$w h a t i s o p t i o n D ?$
Commented by Mikael_786 last updated on 21/Jan/20

$h e l p p l z$
Commented by john santu last updated on 21/Jan/20

$\lim_{x \to 0} [g (x^{2}) \times 2 x] \times \frac{1}{x^{3}} = 1$ $\lim_{x \to 0} \frac{2 g (x^{2})}{x^{2}} = 1 \Rightarrow \lim_{x \to 0} \frac{2 {x g}^{'} (x^{2})}{2 x} = 1$ $g^{'} (0) = 1 .$ $n o w w e w o r k \lim_{x \to 0} \frac{g^{'} (x^{2}) - 5 x}{x - 3} =$ $\frac{g^{'} (0) - 5 \times 0}{0 - 3} = - \frac{1}{3}$
Commented by john santu last updated on 21/Jan/20

$n o t c l e a r o p t i o n D a n d E$
Commented by Mikael_786 last updated on 21/Jan/20

$t h a n k y o u S i r$
Commented by mr W last updated on 21/Jan/20

$w h a t i f e . g . g (x) = \frac{x}{2} (1 + 2 x + 3 x^{2})$ $w h i c h f u l f i l l s t h e c o n d i t i o n b u t$ $g^{'} (0) \neq 1 .$ $o r g (x) = \frac{x \cos x}{2} w h i c h a l s o f u l f i l l s$ $t h e c o n d i t i o n .$
Commented by mr W last updated on 21/Jan/20

$M i k a e l s i r :$ $p l e a s e t r a n s l a t e t h e q u e s t i o n a n d$ $t e l l t h e o p t i o n D, o r E i f e x i s t s!$
Commented by Mikael_786 last updated on 21/Jan/20

Commented by Mikael_786 last updated on 21/Jan/20

$D o p t i o n e x i s t s, b u t I {d o n}^{'} t k n o w t o o .$
Commented by mr W last updated on 21/Jan/20

$i t h i n k t h e a n s w e r c o u l d b e - \frac{1}{6} .$
	Answered by mr W last updated on 21/Jan/20

	$\frac{d \int_{0}^{x^{2}} g (t) d t}{d x} = 2 x g (x^{2})$ $\lim_{x \to 0} \frac{2 x g (x^{2})}{x^{3}} = 1$ $\lim_{x \to 0} \frac{g (x^{2})}{x^{2}} = \frac{1}{2}$ $\Rightarrow g (x^{2}) = \frac{x^{2} h (x)}{2} w i t h \lim_{x \to 0} h (x) = 1$ $\Rightarrow g (x) = \frac{x h (\sqrt{x})}{2}$ $\Rightarrow g^{'} (x) = \frac{h (\sqrt{x})}{2} + \frac{\sqrt{x} h^{'} (\sqrt{x})}{4}$ $\Rightarrow g^{'} (x^{2}) = \frac{h (x)}{2} + \frac{{x h}^{'} (x)}{4}$ $\lim_{x \to 0} g^{'} (x^{2}) = \lim_{x \to 0} \frac{h (x)}{2} + 0 = \frac{1}{2}$ $\lim_{x \to 0} \frac{g^{'} (x^{2}) - 5 x}{x - 3} = \frac{\frac{1}{2} - 0}{0 - 3} = - \frac{1}{6}$
	Commented by john santu last updated on 21/Jan/20

	$\frac{d (g (x^{2}))}{d x} = 2 x g^{'} (x^{2}) s i r$ $w h y \frac{h (x)}{2} + \frac{{x h}^{'} (x)}{4} ?$
	Commented by mr W last updated on 22/Jan/20

	$b y u s i n g t h e f o r m y^{'} y o u s h o u l d b e$ $v e r y c a r e f u l w h a t y o u r e a l l y m e a n,$ $i . e . w i t h r e s p e c t t o w h a t ?$ $a s s u m e g (x) = x^{2} (1 + x) . c a n y o u t e l l m e$ $w h a t g^{'} (x^{2}) i s c o n c r e t e l y i n y o u r e q n .$ $\frac{d (g (x^{2}))}{d x} = 2 x g^{'} (x^{2}) = ?$
	Commented by john santu last updated on 22/Jan/20

	$c h a i n r u l e s i r$ $l e t u = x^{2} \Rightarrow \frac{d u}{d x} = 2 x$ $n o w \frac{d (g (x^{2}))}{d x} = \frac{d (g (u))}{d x}$ $= \frac{d (g (u))}{d u} \times \frac{d u}{d x} = g^{'} (u) \times 2 x$ $t h a t r i g h t ?$
	Commented by mr W last updated on 22/Jan/20

	$y e s!$
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