Let f be a real valued function defined on the interval (−1,1) such that e^(−x) .f(x)=2+∫_0 ^x (√(t^4 +1)) dt ∀x∈(−1,1) and let g be the inverse function of f . Find the value of g′(2).


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Question Number 115260 by bobhans last updated on 24/Sep/20

$L e t f b e a r e a l v a l u e d f u n c t i o n d e f i n e d$ $o n t h e i n t e r v a l (- 1, 1) s u c h t h a t$ $e^{- x} . f (x) = 2 + \underset{0}{\int^{x}} \sqrt{t^{4} + 1} d t \forall x \in (- 1, 1)$ $a n d l e t g b e t h e i n v e r s e f u n c t i o n o f f$ $. F i n d t h e v a l u e o f g^{'} (2) .$
Commented by PRITHWISH SEN 2 last updated on 24/Sep/20

	Answered by john santu last updated on 24/Sep/20

	$D i f f e r e n t i a t i n g g i v e n e q u a t i o n w e$ $g e t e^{- x} . f^{'} (x) - e^{- x} . f (x) = \sqrt{1 + x^{4}}$ $s i n c e (g \circ f) (x) = x a s g i s i n v e r s e o f f$ $\Rightarrow g (f (x)) = x \Rightarrow f^{'} (x) . g^{'} (f (x)) = 1$ $\Rightarrow g^{'} (f (0)) = \frac{1}{f^{'} (0)} \Rightarrow g^{'} (2) = \frac{1}{f^{'} (0)}$ $(h e r e f (0) = 2 o b t a i n e d f r o m$ $g i v e n e q u a t i o n .)$ $p u t x = 0 w e g e t f^{'} (0) = 3 .$ $∴ g^{'} (2) = \frac{1}{3}$
	Answered by mindispower last updated on 25/Sep/20

	$g^{'} (f (x)) = \frac{1}{f^{'} (x)}, f^{'} (x) = e^{x} (2 + \int_{0}^{x} \sqrt{1 + t^{4}}) + \sqrt{x^{4} + 1} e^{x})$ $g^{'} (f (0)) = \frac{1}{f^{'} (0)} = g^{'} (2) = \frac{1}{3}$
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