sinh[ln (x + (√(1 + x^2 ))) ] ≡ A. 2x B. (1/x) C. x^2 D. x


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 71235 by Rio Michael last updated on 13/Oct/19

$s i n h [l n (x + \sqrt{1 + x^{2}})] \equiv$ $A . 2 x$ $B . \frac{1}{x}$ $C . x^{2}$ $D . x$
Commented by mathmax by abdo last updated on 13/Oct/19
$sh{ln(x+(√(1+x^2 )))} =((e^(ln(x+(√(1+x^2 )))) −e^(−ln(x+(√(1+x^2 )))) )/2) =((x+(√(1+x^2 ))−(1/(x+(√(1+x^2 )))))/2) =(((x+(√(1+x^2 )))^2 −1)/(2(x+(√(1+x^2 ))))) =((x^2 +2x(√(1+x^2 )) +1+x^2 −1)/(2(x+(√(1+x^2 ))))) =((2x^2 +2x(√(1+x^2 )))/(2(x+(√(1+x^2 ))))) =((x(x+(√(1+x^2 )))/(x+(√(1+x^2 )))) =x also we can use that ln(x+(√(1+x^2 )))=argsh(x)=sh^(−1) (x)$
$s h {l n (x + \sqrt{1 + x^{2}})} = \frac{e^{l n (x + \sqrt{1 + x^{2}})} - e^{- l n (x + \sqrt{1 + x^{2})}}}{2}$ $= \frac{x + \sqrt{1 + x^{2}} - \frac{1}{x + \sqrt{1 + x^{2}}}}{2} = \frac{{(x + \sqrt{1 + x^{2}})}^{2} - 1}{2 (x + \sqrt{1 + x^{2}})}$ $= \frac{x^{2} + 2 x \sqrt{1 + x^{2}} + 1 + x^{2} - 1}{2 (x + \sqrt{1 + x^{2})}} = \frac{2 x^{2} + 2 x \sqrt{1 + x^{2}}}{2 (x + \sqrt{1 + x^{2}})}$ $= \frac{x (x + \sqrt{1 + x^{2}}}{x + \sqrt{1 + x^{2}}} = x$ $a l s o w e c a n u s e t h a t l n (x + \sqrt{1 + x^{2}}) = a r g s h (x) = {s h}^{- 1} (x)$
	Answered by MJS last updated on 13/Oct/19

	$\sinh α = \frac{1}{2} (e^{α} - e^{- α})$ $\Rightarrow answer is D . x$
	Commented by Rio Michael last updated on 13/Oct/19

	$i {d o n}^{'} t u n d e r s t a n d t h e w o r k i n g s i r$
	Commented by MJS last updated on 13/Oct/19

	$first, sorry I made a typo in the formula$ $\sinh \ln (x + \sqrt{x^{2} + 1}) = \frac{1}{2} (x + \sqrt{x^{2} + 1} - \frac{1}{x + \sqrt{x^{2} + 1}}) =$ $= \frac{1}{2} \times \frac{{(x + \sqrt{x^{2} + 1})}^{2} - 1}{x + \sqrt{x^{2} + 1}} = \frac{1}{2} \times \frac{2 x^{2} + 2 x \sqrt{x^{2} + 1}}{x + \sqrt{x^{2} + 1}} = x$
	Commented by Rio Michael last updated on 13/Oct/19

	$t h a n k s s i r$
	Answered by mr W last updated on 13/Oct/19

	$l e t t = \ln (x + \sqrt{1 + x^{2}})$ $e^{t} = x + \sqrt{1 + x^{2}} . . . (i)$ $e^{- t} = \frac{1}{x + \sqrt{1 + x^{2}}} = - x + \sqrt{1 + x^{2}}$ $- e^{- t} = x - \sqrt{1 + x^{2}} . . . (i i)$ $e^{t} - e^{- t} = 2 x$ $\frac{e^{t} - e^{- t}}{2} = x$ $\Rightarrow \sinh t = x$ $i . e . \sinh [\ln (x + \sqrt{1 + x^{2}})] = x$ $\Rightarrow a n s w e r D$
	Commented by Rio Michael last updated on 13/Oct/19

	$t h a n k s s i r s$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum