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Question Number 123412 by bemath last updated on 25/Nov/20

	Answered by liberty last updated on 25/Nov/20
	$use the formula (f^(−1) )′(a) = (1/(f ′(f^(−1) (a)))) we want to compute the value of (f^(−1) )′(2) = (1/(f ′(f^(−1) (2)))) → { ((let f^(−1) (2) = k then f(k)=2)),((f ′(x)=−((3x^2 )/2)(1+x^3 )^(−(3/2)) )) :} from f(k) = 2 give (1+k^3 )^(−(1/2)) = 2 similar (√(1+k^3 )) = (1/2) ; 1+k^3 = (1/4) k^3 = −(3/4) or k = −((3/4))^(1/3) . Thus (f^(−1) )′(2)=(1/(f ′(−((3/4))^(1/3) ))) (f^(−1) )′(2)= (1/(−(3/2)((9/(16)))^(1/3) (1−(3/4))^(−(3/2)) )) = −((2((1/8)))/(3 ((9/(16)))^(1/3) )) = −(1/(12))(((16)/9))^(1/3)$
	$u s e t h e f o r m u l a {(f^{- 1})}^{'} (a) = \frac{1}{f^{'} (f^{- 1} (a))}$ $w e w a n t t o c o m p u t e t h e v a l u e o f$ ${(f^{- 1})}^{'} (2) = \frac{1}{f^{'} (f^{- 1} (2))}$ $\to {\begin{cases} l e t f^{- 1} (2) = k t h e n f (k) = 2 \\ f^{'} (x) = - \frac{3 x^{2}}{2} {(1 + x^{3})}^{- \frac{3}{2}} \end{cases}$ $f r o m f (k) = 2 g i v e {(1 + k^{3})}^{- \frac{1}{2}} = 2$ $s i m i l a r \sqrt{1 + k^{3}} = \frac{1}{2}; 1 + k^{3} = \frac{1}{4}$ $k^{3} = - \frac{3}{4} o r k = - \sqrt[3]{\frac{3}{4}} . T h u s {(f^{- 1})}^{'} (2) = \frac{1}{f^{'} (- \sqrt[3]{\frac{3}{4}})}$ ${(f^{- 1})}^{'} (2) = \frac{1}{- \frac{3}{2} \sqrt[3]{\frac{9}{16}} {(1 - \frac{3}{4})}^{- \frac{3}{2}}}$ $= - \frac{2 (\frac{1}{8})}{3 \sqrt[3]{\frac{9}{16}}} = - \frac{1}{12} \sqrt[3]{\frac{16}{9}}$
	Commented by Bird last updated on 25/Nov/20

	$c o r r e c t!$
	Answered by TANMAY PANACEA last updated on 25/Nov/20

	$y = f (x) a n d (a, b) l i e s o n y = f (x)$ $b = f (a)$ $f o r m u l a [f^{- 1}] (b) = \frac{1}{f^{^{'}} (a)}$ $h e r e b = 2$ $2 = {(1 + a^{3})}^{\frac{- 1}{2}}$ $2 = \frac{1}{{(1 + a^{3})}^{\frac{1}{2}}}$ $4 = \frac{1}{1 + a^{3}} \to 4 + 4 a^{3} = 1$ $4 a^{3} = - 3$ $a = {(\frac{- 3}{4})}^{\frac{1}{3}}$ $f (x) = {(1 + x^{3})}^{\frac{- 1}{2}}$ $f^{^{'}} (x) = \frac{- 1}{2} \times {(1 + x^{3})}^{\frac{- 3}{2}} \times 3 x^{2}$ $r e q u i r e d a n s w d r$ $= \frac{1}{{(\frac{- 3 x^{2} {(1 + x^{3})}^{\frac{- 3}{2}}}{2})}_{x = a = {(\frac{- 3}{4})}^{\frac{1}{3}}}}$ $P l s c a l v u l a t d$
	Answered by Bird last updated on 25/Nov/20

	${(f^{- 1})}^{^{'}} (2) = \frac{1}{f^{^{'}} (f^{- 1} (2))}$ $f^{- 1} (x) = y \Rightarrow x = f (y) = {(1 + y^{3})}^{- \frac{1}{2}}$ $\Rightarrow x^{2} = \frac{1}{1 + y^{3}} \Rightarrow x^{2} + x^{2} y^{3} = 1 \Rightarrow$ $x^{2} y^{3} = 1 - x^{2} \Rightarrow y^{3} = \frac{1 - x^{2}}{x^{2}}$ $= \frac{1}{x^{2}} - 1 \Rightarrow y = {(x^{- 2} - 1)}^{\frac{1}{3}} \Rightarrow$ $f^{- 1} (x) = {(x^{- 2} - 1)}^{\frac{1}{3}} \Rightarrow$ ${(f^{- 1})}^{^{'}} (x) = \frac{1}{3} (- 2 x^{- 3}) {(x^{- 2} - 1)}^{- \frac{2}{3}}$ $\Rightarrow {(f^{- 1})}^{^{'}} (2) = - \frac{2}{3} . 2^{- 3} {(2^{- 2} - 1)}^{- \frac{2}{3}}$ $= - \frac{1}{3.4} {(\frac{1}{4} - 1)}^{- \frac{2}{3}}$ $= - \frac{1}{12} {(- \frac{3}{4})}^{- \frac{2}{3}}$ $= - \frac{1}{12} \times {(- \frac{4}{3})}^{\frac{2}{3}}$ $= \frac{- 1}{12} (^{3} \sqrt{\frac{16}{9}})$
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