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I was looking at a question that asks for the derivative of $\arcsin (\frac {x+1}{x-1}) $.

The solution starts by saying $y = \frac{x+1}{x-1}$, so $1-y^2= \frac{4x}{(x+1)^2}$ and $\frac{1}{\sqrt{1-y^2}}$ and thus $\frac{x+1}{2\sqrt{x}} $

However, I don't see why $1-y^2= \frac{4x}{(x+1)^2} $

Michael Hardy
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Haim
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  • $1-y^2=1-\left(\dfrac{x-1}{x+1}\right)^2=1-\dfrac{x^2-2x+1}{x^2+2x+1}=\dfrac{(x^2+2x+1)-(x^2-2x+1)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}=\dfrac{4x}{(x+1)^2}$ – Kenny Lau Jun 23 '16 at 15:54
  • I think there is a typo, it must be $1-y^2=\dfrac{4x}{(x-1)^2}$ as you have defined $y = \dfrac{x+1}{x-1}$. As for the rest, Kenny lau's calculus shows you the answer. – ysearka Jun 23 '16 at 16:01
  • @ysearka The title doesn't even match the first line of the question, so I just picked whichever I like. – Kenny Lau Jun 23 '16 at 16:03
  • @Kenny Lau, Yes I was saying there's a typo in the question, as it's not coherent. – ysearka Jun 23 '16 at 16:07

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$1-y^2=1-\left(\dfrac{x-1}{x+1}\right)^2=1-\dfrac{x^2-2x+1}{x^2+2x+1}=\dfrac{(x^2+2x+1)-(x^2-2x+1)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}=\dfrac{4x}{(x+1)^2}$

Kenny Lau
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