Question

If $y=\cos \left(\frac{1}{2}{\cos }^{-1}x\right)$, then:$\frac{dy}{dx}=$

Answer 1

Let $t={\cos }^{-1}x$

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

$y=\cos \frac{t}{2}$

$\Rightarrow \frac{dy}{dt}=\frac{-1}{2}\sin \frac{t}{2}$

$\therefore \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{-1}{2}\sin \frac{t}{2}\times \frac{-1}{\sqrt{1-x^2}}$

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

$=\frac{1}{2\sqrt{1-x^2}}\sin \left(\frac{\left({\cos }^{-1}x\right)}{2}\right)$ where $t={\cos }^{-1}x$