compute the derivative
G(s) =[tex] cos^{-1} (s^{-1}) [/tex]

I've gotten up to [tex]1 / ( \sqrt{1-(1/ s^{2} )[/tex]
the actual answer says there should be an [tex] s^{2} [/tex] outside of the radical and I'm not sure how to get that.

Respuesta :

[tex]\bf g(s)=cos^{-1}(s^{-1})\\\\ -----------------------------\\\\ \cfrac{dg}{ds}=\cfrac{-1}{\sqrt{1-s^{-2}}}\cdot -s^{-2}\implies \cfrac{dg}{ds}=\cfrac{-1}{\sqrt{1-s^{-2}}}\cdot\cfrac{-1}{s^2}\impliedby \begin{array}{llll} using\ the\\ chain-rule \end{array} \\\\\\ \cfrac{dg}{ds}=\cfrac{-1}{\sqrt{1-\frac{1}{s^2}}}\cdot\cfrac{-1}{s^2}\implies \boxed{\cfrac{dg}{ds}=\cfrac{1}{s^2\sqrt{1-\frac{1}{s^2}}}}[/tex]
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