Question

Find $\frac{dy}{dx}$

$y={\cos }^{-1}\left(\sqrt{\frac{1+x}{2}}\right)$

Answer 1

$Letx=cos\theta or\theta =co{s}^{-1}xtheny=co{s}^{-1}\left(\sqrt{\left(\frac{1+cos\theta }{2}\right)}\right)=co{s}^{-1}\left(\sqrt{\left(\frac{1+\left(2co{s}^2\right(\frac{\theta }{2})-1)}{2}\right)}\right)=co{s}^{-1}\left(cos\right(\frac{\theta }{2}\left)\right)=\left(\frac{\theta }{2}\right)=\left(\frac{1}{2}\right)co{s}^{-1}x\therefore \left(\frac{dy}{dx}\right)=\left(\frac{1}{2}\right)\times \left(\frac{-1}{\sqrt{1-x^2}}\right)=\left(\frac{-1}{2\sqrt{1-x^2}}\right)$