Question

$y=co{s}^{-1}\left(\sqrt{\frac{1+x}{2}}\right)$ find $\frac{dy}{dx}$

Answer 1

We have,

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

Then $\frac{dy}{dx}=?$

Suppose that,

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

On differentiating with respect to x and we get,

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

$=-\frac{1}{\sqrt{1-{\left(\sqrt{\frac{1+x}{2}}\right)}^2}}\frac{d}{dx}\left(\sqrt{\frac{1+x}{2}}\right)\because \frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$

$=-\frac{1}{\sqrt{1-\frac{1+x}{2}}}\frac{1}{2\sqrt{\frac{1+x}{2}}}\frac{d}{dx}\left(\frac{1+x}{2}\right)$

$=-\frac{1}{2\sqrt{\frac{2-1-x}{2}}}\frac{1}{\sqrt{\frac{1+x}{2}}}\frac{d}{dx}\left(\frac{1+x}{2}\right)$

$=-\frac{1}{2}\frac{1}{\sqrt{\frac{1-x}{2}}}\frac{1}{\sqrt{\frac{1+x}{2}}}\frac{1}{2}(0+1)$

$=-\frac{1}{4}\frac{1}{\sqrt{\left(\frac{1-x}{2}\right)\left(\frac{1+x}{2}\right)}}$

$=-\frac{1}{4}\frac{1}{\frac{\sqrt{1-x^2}}{2}}$

$=-\frac{1}{2}\frac{1}{\sqrt{1-x^2}}$

Hence, this is the answer.