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Question Number 145164 by mim24 last updated on 02/Jul/21

	Answered by mathmax by abdo last updated on 02/Jul/21

	$this question is solved see the platform$
	Commented by mim24 last updated on 02/Jul/21

	$y o u m a d$
	Commented by mathmax by abdo last updated on 03/Jul/21

	$ni mad ni hmad i am mathmax at this forum and its seems that$ $you crazy . . .!$
	Answered by EDWIN88 last updated on 03/Jul/21

	$l e t \frac{x - x^{- 1}}{x + x^{- 1}} = \cos y$ $\Rightarrow \frac{x^{2} - 1}{x^{2} + 1} = \cos y$ $\Rightarrow \frac{x^{2} + 1 - 2}{x^{2} + 1} = \cos y$ $\Rightarrow 1 - 2 {(x^{2} + 1)}^{- 1} = \cos y$ $\Rightarrow 2.2 x . {(x^{2} + 1)}^{- 2} = - \sin y . \frac{d y}{d x}$ $\Rightarrow \frac{- 4 x}{{(x^{2} + 1)}^{2} . \sin y} = \frac{d y}{d x}$ $\Rightarrow \frac{- 4 x}{{(x^{2} + 1)}^{2} \sqrt{1 - {(\frac{x^{2} - 1}{x^{2} + 1})}^{2}}} = \frac{d y}{d x}$ $\Rightarrow \frac{d y}{d x} = \frac{- 4 x}{(x^{2} + 1) \sqrt{{(x^{2} + 1)}^{2} - {(x^{2} - 1)}^{2}}}$ $\frac{d y}{d x} = \frac{- 4 x}{(x^{2} + 1) \sqrt{4 x^{2}}} = \frac{- 4 x}{(x^{2} + 1) ∣ 2 x ∣}$
	Answered by puissant last updated on 03/Jul/21
	$arccos(((x−(1/x))/(x+(1/x))))=arccos(((x^2 −1)/(x^2 +1))) (d/dx)(arccos(((x^2 −1)/(x^2 +1))))=(((d/dx)(((x^2 −1)/(x^2 +1))))/( (√(1−(((x^2 −1)/(x^2 +1)))^2 )))) =(((2x(x^2 +1)−2x(x^2 −1))/((x^2 +1)^2 ))/( (√(((x^2 +1)^2 −(x^2 −1)^2 )/((x^2 +1)^2 ))))) = (((4x)/((x^2 +1)))/( (√((x^2 +1−x^2 +1)(x^2 +1+x^2 −1))))) =((4x)/((x^2 +1)))×(1/(2∣x∣)) ⇒ (d/dx)(cos^(−1) (((x−x^(−1) )/(x+x^(−1) ))))={_((2/(x^2 +1)) , x>0) ^(((−2)/(x^2 +1)) , x<0)$
	$\arccos (\frac{x - \frac{1}{x}}{x + \frac{1}{x}}) = \arccos (\frac{x^{2} - 1}{x^{2} + 1})$ $\frac{d}{dx} (\arccos (\frac{x^{2} - 1}{x^{2} + 1})) = \frac{\frac{d}{dx} (\frac{x^{2} - 1}{x^{2} + 1})}{\sqrt{1 - {(\frac{x^{2} - 1}{x^{2} + 1})}^{2}}}$ $= \frac{\frac{2 x (x^{2} + 1) - 2 x (x^{2} - 1)}{{(x^{2} + 1)}^{2}}}{\sqrt{\frac{{(x^{2} + 1)}^{2} - {(x^{2} - 1)}^{2}}{{(x^{2} + 1)}^{2}}}} = \frac{\frac{4 x}{(x^{2} + 1)}}{\sqrt{(x^{2} + 1 - x^{2} + 1) (x^{2} + 1 + x^{2} - 1)}}$ $= \frac{4 x}{(x^{2} + 1)} \times \frac{1}{2 ∣ x ∣}$ $\Rightarrow \frac{d}{dx} (\cos^{- 1} (\frac{x - x^{- 1}}{x + x^{- 1}})) = {_{\frac{2}{x^{2} + 1}, x > 0}^{\frac{- 2}{x^{2} + 1}, x < 0}$
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