Question

If $y={\cos }^{-1}(\sqrt{\frac{1+x}{2})}$ .Find $\frac{dy}{dx}$ .

Answer 1

We have,

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

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

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

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

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

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

$\frac{dy}{dx}=-\frac{2}{1+x}$

Hence, this is the answer.