pierdoodsu

2021-12-30

Derivative of ${\mathrm{cos}}^{-1}\sqrt{\frac{1+x}{2}}$ using substitution

John Koga

Setting $x=\mathrm{cos}2\theta$ and remembering that $\mathrm{cos}2\theta =2\mathrm{cos}2\theta -1$ leads to:
${\mathrm{cos}}^{-1}\sqrt{\frac{1+x}{2}}={\mathrm{cos}}^{-1}\sqrt{\frac{1+\mathrm{cos}2\theta }{2}}$
$={\mathrm{cos}}^{-1}\sqrt{\frac{1+2{\mathrm{cos}}^{2}\theta -1}{2}}$
$=\theta$
And so one can use:
$\frac{d}{dx}{\mathrm{cos}}^{-1}\sqrt{\frac{1+x}{2}}=\frac{d\theta }{dx}=\frac{1}{\frac{dx}{d\theta }}$

Thomas Nickerson

Let $x=\mathrm{cos}\left(2\theta \right)$ then
${\mathrm{cos}}^{-1}\sqrt{\frac{1+x}{2}}={\mathrm{cos}}^{-1}\left(\mathrm{cos}\theta \right)=\theta$
and
$\frac{dx}{d\theta }=-2\mathrm{sin}\left(2\theta \right)$
$\frac{d}{dx}\left({\mathrm{cos}}^{-1}\sqrt{\frac{1+x}{2}}\right)=-\frac{12}{\frac{1}{\mathrm{sin}\left(2\theta \right)}}=-\frac{12}{\frac{1}{\sqrt{1-{x}^{2}}}}$

nick1337

I think the natural way should be:
Let ${\mathrm{cos}}^{-1}\sqrt{\frac{1+x}{2}}=y$
$⇒0\le y\le \frac{\pi }{2}$ for real $1\ge x\ge -1$
$⇒x=\dots =\mathrm{cos}2y$
and consequently

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